Integral

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Introduction

Solution 👉 Click Here

Derivative

1. instanteneous rate of change
2. slope of tangent

Anti-Derivative Let f(x) be a continuous function defined in an open interval (a,b), then a function F(x)+c is called an antiderivative of f(x) if
\( \frac{dy}{dx}[F(x)+c]=f(x)\)

1. reverse process of derivative
   If \( \frac{dy}{dx}F(x)=f(x)\) , then \( F(x)\) is called an anti-derivative of \( f(x)\)
2. family of functions
   the anti-drivative of f(x) always infine in number, they are F(x)+c
   For example, the anti-derivative of \(3x^2\) is \(x^3+c\), it can be
   \(x^3+1,x^3+2,x^3+3,x^3+4,x^3+\pi,x^3+\sqrt{2},x^3+e,x^3+0.3,\cdots\)

Integral

1. a limiting process, defined as the limit of sum of rectangles under f(x)
2. sum of products, adding peice of multiplication followed by addition: an infinite sum of extremely narrow rectangles, each rectangle having a height f(x) and a width dx; the differential of x, graphical area enclosed by the function and the interval boundaries

Mathematically

Given a function f(x), the anti-derivative of f(x) is defined as a new function F(x) given by
F'(x)=f(x)

Blue curves are anti-derivatives

Blue curves are anti-derivatives

Derivative and anti-derivative
1. The derivative of f(x) is F(x): the slope of f(x) is represented by F(x)
2. The integral of F(x) is f(x): the mass of F(x) is represented by f(x)

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Historical Evidence

1. method of exhaustion in Greek (Eudoxus-300BC)
2. similar method in China (Liu Hui-200AD)
3. first proof of the fundamental theorem of calculus(Barrow-1600AD)
4. independent discovery of the fundamental theorem of calculus(Leibniz and Newton-1600AD)
5. notion were used [Lebniz-∫, Fourier-with limits]

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Integral as Riemann Sum

Solution 👉 Click Here

The integral is defined in terms of Riemann sums of functions with respect to tagged partitions of an interval. A tagged partition of a closed interval [a, b] on the real line is a finite sequence given by
\( a=x_0 \leq x_1 \leq x_2 \leq x_3 \leq ... \leq x_n=b\)
This partitions the interval [a, b] into n sub-intervals\( [x_{i−1}, x_i] \) of length \(\Delta x_i\) indexed by i.
A Riemann sum of a function f(x) with respect to such a tagged partition is defined as
\( \int_a^b f(x)dx= \displaystyle \sum _{i=1}^n f(x_i) \Delta x_i\)
where \(x_i\) can be any number inside the \( [x_{i−1}, x_i] \)

Upper Sum

Lower Sum

Left Sum

Right Sum

Middle Sum

Graph of Riemann Sum

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The Integral: Example

Use rectangles to estimate the area under the parabola \(y=-0.5x^2+2x\) from 0 to 4.
Solution
The area is
Area=\(\displaystyle \lim_{n \to \infty} \sum_{i=1}^n f(x_i) \Delta x\)
where
\(f(x_i)\) is the length of the rectangle
\(\Delta x\) is the breath of the rectangle
The length of the rectangle can be found from the

1. upper point
2. lower point
3. left point
4. right point
5. middle point

Example

Evaluate \(\displaystyle \int_0^4(x^2)dx\)
Solution
With n subintervals, we have
\(\Delta x=\frac{b-a}{n}=\frac{4}{n}\)
Using the right end points, we get

\(\displaystyle \int_0^4(x^2)dx\)	\( \displaystyle = \lim_{n \to \infty} \sum_{i=1}^n f \left ( \frac{4i}{n} \right) (\Delta x)\)
	\(\displaystyle= \lim_{n \to \infty} \sum_{i=1}^n \left ( \frac{4i}{n} \right)^2 \left ( \frac{4}{n} \right ) \) \(\displaystyle= \lim_{n \to \infty} \frac{64}{n^3} \sum_{i=1}^n (i^2) \) \(\displaystyle= \lim_{n \to \infty} \frac{64}{n^3} \frac{n(n+1)(2n+1)}{6} \) \(\displaystyle= \lim_{n \to \infty} 64 \frac{1(1+1/n)(2+1/n)}{6} \) \(= 64 \frac{1(1+0)(2+0)}{6} \) \( =\frac{64}{3} \) \( = 21.33\)

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Fundamental Theorem of Calculus

Solution 👉 Click Here

Before, we start Fundamental Theorem of Calculus(FTC), we will prove Mean Value theorem of Integral.

Mean Value theorem of Integral

Let f(x) : [a, b] → R be a continuous function. Then there exists c in [a, b] such that
\( \displaystyle \int _{a}^{b}f(x)dx=f(c)(b-a)\)
Since the mean value of f(x) on [a, b] is defined as \( \frac{1}{b-a} \int _a^b f(x)dx=f(c)\)
We can interpret the conclusion as f(x) achieves its mean value at some c in (a, b)
Thus
Let f(t) is continuous function on [x,x+h], then there exists c in [x,x+h] such that
\( \frac{1}{(x+h)-x} \int_{x}^{x+h} f(t)dt=f(c)\)

or \( \frac{1}{h} \int_{x}^{x+h} f(t)dt=f(c)\)

Now
Fundamental Theorem of Calculus,shows that
\( \frac{d}{dx} \left [ \int_a^b f(x)dx \right ]=f(x) \)

The fundamental theorem of calculus links the concept of differentiation and integration. It justifies the procedure by computing the integral using antiderivative
Thus, gives us a way to evaluate definite integrals without using Riemann sums or calculating areas.

Fundamental Theorem (First Part)

This theorem ensures that any integrable function has an antiderivative. Specifically, it guarantees the existence of antiderivative. It is also called theorem for Integrals and Antiderivatives

Statement
Let f(x) is continuous on [a, b] and F(x) is defined by
\( \displaystyle F(x)=\int _{a}^{x}f(t)dt\) for all x in [a, b]
Then
\( \displaystyle F'(x)=f(x)\) for all x in (a, b)

Proof
By definition, we choose a base point such that
\( \displaystyle F(x)=\int _a^x f(t)dt\) (1)
\( \displaystyle F(x +h)=\int _{a}^{x +h}f(t)dt\) (2)
Applying the definition of the derivative, we have
\(\displaystyle F'(x )=\displaystyle \lim_{h \to 0} \frac{F(x+h)-F(x)}{h} \)
or \(\displaystyle F'(x )=\displaystyle \lim_{h \to 0} \frac{1}{h} \left [ F(x+h)-F(x) \right ] \)
or \(\displaystyle F'(x )=\displaystyle \lim_{h \to 0} \frac{1}{h} \left [ \int _{a}^{x +h}f(t)dt -\int _{a}^{x}f(t)dt \right ] \)Using (1) and (2)
or \(\displaystyle F'(x )=\displaystyle \lim_{h \to 0} \frac{1}{h} \left [ \int _{x}^{x +h}f(t)dt \right ] \)(3)
By Mean Value Theorem for Integrals in (3), there is some number c in [x,x+h] such that
\(\displaystyle F'(x )=\displaystyle \lim_{h \to 0} f(c) \)
Since c is between x and x + h, c approaches x as h approaches zero.
Thus
\(\displaystyle F'(x )=f(x) \)
This completes the proof.

Fundamental Theorem (Second Part)

The Fundamental Theorem of Calculus, Part 2 is also known as the evaluation theorem. It states that if we can find an antiderivative for the integrand, then we can evaluate the definite integral by evaluating the antiderivative at the endpoints of the interval and subtracting.

Statement
Let f(x) is integrable on [a, b], and f(x) admit its antiderivative F(x), then
\(\displaystyle F(b)-F(a)=\int_{a}^{b}f(x)dx\)

Proof
Let \( g(x)=\int_a^x f(t)dt\), then by The Fundamental Theorem of Calculus Part 1, there exist g(x), which is an antiderivative of f(x), that is
\(g′(x)=f(x)\)
Also let, F(x) be antiderivative of f(x), then it follows that
\(F(x)=g(x)+c\) where C is a constant
Note: F(x) and g(x) differ only by a constant
Now
\( g(a)=\int_a^a f(t)dt=0\)
Next,
\(F(b)−F(a)=[g(b)+c]−[g(a)+c] \)
or \(F(b)−F(a)=g(b)−g(a) \)
or \(F(b)−F(a)=g(b)\)
or \(F(b)−F(a)=\int_a^b f(t)dt\)
or \( F(b)−F(a)=\int_a^b f(x)dx\)

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Relation between Derivative, Anti-Derivative and Integral

Solution 👉 Click Here

In the definite form, fundamental theorem of calculus connects differentiation with the definite integral as follows.
If f(x) is a continuous real-valued function defined on a closed interval [a, b], then integral of f over [a, b] is given by
\( \int_a^b\) f(x)dx=F(b)-F(a)

For example,
मानौ, y=x एउटा function हो, जसको कुनै एउटा बिन्दुले X-axis सँग बनाउने Area = \( \frac{1}{2}x^2\) छ।

यहाँ,
Integral of x is \( \frac{1}{2}x^2\) हुन्छ ।
Derivative of \( \frac{1}{2}x^2\) is x हुन्छ ।
त्यसैले, Integral is inverse process of differentiation भनिएको हो।

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Derivative, Anti-Derivative and Integral

Derivative, Anti-Derivative and Integral are close due to Fundamental Theorem of Calculus (FTC). Before this theorem, anti-derivative and integral were not thought of as related to each other.

1. Derivative and Anti-Derivative

   Derivative and Anti-derivative are very close due to Fundamental Theorem of Calculus (FTC, I).
   Derivative is an infinitesimal increment of change (difference), which is
   • a quotient of differences(changes)
   • a graphical slope (its “rise over run”),and slope of tangent line is the derivative of the function at that point
   • specifid at a single point along the function where the slope is to be calculated
   Anti-derivative of f(x) is F(x)+c such that [F(x)+c]'=f(x).
   • a family of functions whose slopes are same
   • The process of executing antiderivative is called antidifferentiation
   Both are part of differential calculus, calculus of difference/change, is about how fast a dependent variable changes as independent variable changes

2. Anti-Derivative and Integral

   Anti-Derivative and Integral are very close due to Fundamental Theorem of Calculus (FTC, II).
   Anti-derivative of f(x) is F(x)+c such that [F(x)+c]'=f(x).
   • a family of functions whose slopes are same
   • The process of executing antiderivative is called antidifferentiation
   • differential calculus, calculus of difference/change, is about how fast a dependent variable changes as independent variable changes
   Integral of f(x) is defined by a limiting process, defined as the limit of sum of rectangles under f(x)
   • sum of products, adding peice of multiplication followed by addition
   • an infinite sum of extremely narrow rectangles, each rectangle having a height f(x) and a width dx; the differential of x, graphical area enclosed by the function and the interval boundaries
   • The process of executing integral is called integration
   • Integral calculus is the calculus of accumulation, is the sum of functions
   There is a very small difference between integral f(x) and anti-derivative F(x), let \(f(x) = x^2\), then,
   1. definite integral is \(\int _𝑎^𝑏 𝑥^2 𝑑𝑥=\frac{1}{3} (b^3 – a^3)=F(b) – F(a) \).
   2. indefinite integral is \(\int 𝑥^2 𝑑𝑥=\frac{x^3}{3}+c \) such that \(\frac{d}{dx} \left (\frac{x^3}{3} \right )=x^2\).

3. Derivative and Integral

   Derivative and Integral are very close due to Fundamental Theorem of Calculus (FTC, I,II).
   Derivative is an infinitesimal increment of change (difference), which is
   • a quotient of differences (changes)
   • a graphical slope (its “rise over run”) ,and slope of tangent line is the derivative of the function at that point
   • specifid at a single point along the function where the slope is to be calculated
   • differential calculus, calculus of difference/change, is about how fast a dependent variable changes as independent variable changes
   Integral of f(x) is defined by a limiting process, defined as the limit of sum of rectangles under f(x)
   • sum of products, adding peice of multiplication followed by addition
   • an infinite sum of extremely narrow rectangles, each rectangle having a height f(x) and a width dx; the differential of x, graphical area enclosed by the function and the interval boundaries
   • The process of executing integral is called integration
   • Integral calculus is the calculus of accumulation, is the sum of functions
   Just as division and multiplication are inverse mathematical functions, differentiation and integration are also inverse mathematical functions. differentiation always divides while integration always multiplies.

Type of Integral

1. Integration as inverse operator
2. Substitutio/Replacement: Derivative in numirator \( \int f^n.f' dx\) or \( \int \frac{f' dx}{f^n} \)
3. By Evaluation
4. By Parts
5. Of the form: \([f(x)+f'(x)] e^x dx\)
6. Partial Fraction

Mean value theorem

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In mathematics, the mean value theorem (or Lagrange theorem) states, roughly, that for a given planar arc between two endpoints, there is at least one point at which the tangent to the arc is parallel to the secant through its endpoints.
More precisely, the theorem states that
if a function f(x) is
(i)continuous in the closed interval [a, b]
(ii) differentiable in the open interval (a, b).
Then, there exists at least one number c∈(a, b) such that
\(f '(c) = \frac{f(b)-f(a)}{b-a}\)

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Geometrical Meaning of Mean Value Theorem

Let f(x) be a continuous function in an interval [a,b] with A=f(a) and B=f(b). Then Mean Value Theorem asserts that there exist at least one point c lying between A and B such that the tangent at that point is parallel to the chord AB as shown in figure below.

There may exist more than one points \(c_1, c_2\) and may be more between A and B, so that the tangents at those points are parallel to chord AB shown in figure.

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Proof of Mean value theorem

Statement
if a function f(x) is
(i)continuous in the closed interval [a, b]
(ii) differentiable in the open interval (a, b).
Then, there exists at least one number c∈(a, b) such that
\(f '(c) = \frac{f(b)-f(a)}{b-a}\)
Proof

Let P(x,y) be any arbitrary point on the line AB.
Then equation of the line AB is given by
\(y-f(a)=\frac{f(b)-f(a)}{b-a} (x-a)\)
or \(y=f(a)+\frac{f(b)-f(a)}{b-a} (x-a)\)
Take a point C, which lies on the function, the value of segment AC, which is perpendicular to x-axis is
AC=f(x)
Now, the value of the segment PC is given by
PC=f(x)- y

Let g(x)=PC is a new function, then
g(x)=f(x)-y
The function g(x) is defined by
\(g(x) = f(x)-\) \(f(a)-\frac{f(b)-f(a)}{b-a} (x-a)\)
Here
g(a)=g(b)
Further g(x) has the same continuity and differentiability properties as f(x)
Thus we can apply Rolle’s theorem to g(x) finding c ∈(a, b), where g'(c) = 0.
We immediately get that
\( g'(c)=0\)
or \( f'(c)-\frac{f(b)-f(a)}{b-a}=0\)
or \(f'(c) = \frac{f(b)-f(a)}{b-a}\)
This completes the proof.

Mean Value Theorem:Solved Examples

1. Which of the following function satisfy the hypothesis of Mean Value theorem, give reason.
   1. \( f(x)=x^2\) over [0,2]

      Solution 👉 Click Here
   2. \( f(x)=x^2-5x+1\) over [0,3]
   3. \( f(x)=x^2-2x-8\) over [-1,3]
   4. \( f(x)=x^3+3x^2-24x\) over [1,4]
   5. \( f(x)=4x^3-8x^2+7x-2\) over [2,5]
   6. \( f(x)=x^3-8x-5\) over [1,4]
   7. \( f(x)=2^x\) over [0,3]
   8. \( f(x)=x^{2/3}\) over [-1,8]
   9. \( f(x)=x^{4/5}\) over [0,1]
   10. \( f(x)=\frac{x+2}{x}\) over [1,2]
   11. \( f(x)=\frac{2}{x}\) over [-1,1]
   12. \( f(x)=5-\frac{4}{x}\) over [1,4]
   13. \( f(x)=x \sqrt{1-x}\) over [0,1]
   14. \( f(x)=2 \sin x+3 \cos x\) over \([0,2 \pi]\)
   15. \( f(x)= \begin{cases} \frac{\sin x}{x} & \pi \le x < 0 \\ 0 & x=0 \end{cases} \)
2. For what value of a,m, and b does the function satisfy Mean value theorem in the interval [0,2]
   \( f(x)= \begin{cases} 3 & x =0\\ -x^2+3x+a & 0 < x < 1\\ mx+b &1 \le x \le 2 \end{cases} \)

Rolle's Theorem

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Statement: If f(x) is
(i) continuous in the closed interval [a, b];
(ii) differentiable in the open interval (a, b).
(iii) f(a)=f(b)
Then, there exists at least one number \(c\in (a, b)\) such that f '(c) = 0.

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Geometrical Meaning of Roll’s Theorem

Let f(x) be a continuous function in an interval [a,b] with A=f(a) and B=f(b).
Then Role’s Theorem asserts that there exist at least one point c lying between A and B such that the tangent at that point is parallel to x-axis.
There may exist more than one points \(c_1, c_2\) and may be more between A and B, so that the tangents at those points are parallel to x-axis

Solution 👉 Click Here

This function \(f(x)=-x^2+4x+5\) meets all the criteria for Rolle's Theorem in the interval [-2,6]. It's continuous on the interval [-2, 6], differentiable in the interval (-2, 6), and f(-2) equals f(6), both equaling to each other.
As expected, there exists at least one point where the derivative is zero. In this case, it happens at x = 2, where the tangent line is horizontal.

This function \(f(x)=(x+4)^2-4(x+4)+4\) meets all the criteria for Rolle's Theorem in the interval [-6,2]. It's continuous on the interval [-6,2], differentiable in the interval (-6,2), and f(-6) equals f(2), both equaling to each other.
As expected, there exists at least one point where the derivative is zero. In this case, it happens at x = -2, where the tangent line is horizontal.

This function \(f(x)=-(x-2)^2+4(x-2)+5\) meets all the criteria for Rolle's Theorem in the interval [1,7]. It's continuous on the interval [1,7], differentiable in the interval (1,7), and f(1) equals f(7), both equaling to each other.
As expected, there exists at least one point where the derivative is zero. In this case, it happens at x = 4, where the tangent line is horizontal.

This function \(f(x)=x^3-4x\) meets all the criteria for Rolle's Theorem in the interval [-2,2]. It's continuous on the interval [-2,2], differentiable in the interval (-2,2), and f(-2) equals f(2), both equaling to each other.
As expected, there exists at least one point where the derivative is zero. In this case, it happens at x = -1.15 and x=1.15, where the tangent line is horizontal.

All three conditions of Rolle’s theorem are necessary for the theorem to be true

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The condition where Rolle’s theorem fails

The condition where Rolle’s theorem is NOT applicable, of the Rolle’s theorem fails.

Discontinuous	Discontinuous	Not Differentiable

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Rolle’s theorem fails: Counter Examples

Solution 👉 Click Here

1. Counter Example 1

   Consider f(x)={x}, a fractional part function on the closed interval [0,1].
   The derivative of the function on the open interval (0,1) is everywhere equal to 1.
   Eventhough, the Rolle’s theorem fails because the function f(x)={x} has a discontinuity at x=0 and x=1
   that is
   the function f(x)={x} is not continuous everywhere on the closed interval [0,1])

2. Counter Example 2

   Consider f(x)=|x|, an absolute value function on the closed interval [-1,1].
   This function does not have derivative at x=0.The Rolle’s theorem fails here because f(x) is not differentiable over the whole interval (-1,1).

3. Counter Example 3

   Consider f(x)=x, a function on the closed interval [0,1].
   Though f(x) is continuous on the closed interval [-1,1], differentiable in the open interval (0,1), it does not satisfy third condition; i.e., f(0)≠f(1).Thus , the Rolle’s theorem fails .

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Rolle’s theorem :History

The Roll's theorem says that, if a continuous curve passes through the same y-value (such as the x-axis) twice and has a unique tangent line (derivative) at every point of the interval, then somewhere between the endpoints it has a tangent parallel to the x-axis. The theorem was proved in 1691 by the French mathematician Michel Rolle, though it was stated without a modern formal proof in the 12th century by the Indian mathematician Bhaskara II.

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Rolle’s theorem :Proof

Statement: If f(x) is
(i) continuous in the closed interval [a, b];
(ii) differentiable in the open interval (a, b).
(iii) f(a)=f(b)
Then, there exists at least one number \(c\in (a, b)\) such that f '(c) = 0.

Proof

If f(x) is constant then
f'(x) (c) = 0 for all c∈(a, b)
Suppose there exists x ∈ (a, b) such that
f(x) > f(a)
(A similar argument can be given if f(x) < f(a)).
Then there exists c ∈ (a, b) such that f(c) is a maximum.
Since, f(x) has a local maximum at c ∈(a, b), For small (enough) h,
\(f(c + h) \le f(c)\).
If h > 0 then
\( \frac{f(c + h)-f(c)}{h} \le 0\).
Similarly,
If h < 0, then
\(\frac{f(c + h)-f(c)}{h} \ge 0\).
By elementary properties of the limit, it follows that
f'(c) = 0.

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Rolle’s theorem: Solved Examples

1. Veryfy that the function satisfy the three hypothesis of Rolle's theorem, and find all numbers c that satisfy the conclusion of Rolle's theorem
   1. \( f(x)=x^2+2x\) over [-2,0]

      Solution 👉 Click Here

      Since f(x) is a polynomial, it is continuous and differentiable everywhere.
      In addition
      f(-2)=0=f(0)
      Therefore, f(x) satisfies three hypothesis of Rolle's theorem. Since, given function is
      \( f(x)=x^2+2x\)
      Differentiating the function w.r. to x, we get
      \( f'(x) = 2x+2\)
      or \( f'(x) = 2(x+1)\)
      According to Rolle's theorem, there exists at least one value \( c\in (-2,0)\) such that f′(c)=0
      So,
      \( f′(c)=0\)
      We see that
      \(2(c+1)=0 \)
      or c=-1
      The figure is shown below
   2. \( f(x)=x^3-4x\) over [-2,2]

      Solution 👉 Click Here

      Since f(x) is a polynomial, it is continuous and differentiable everywhere.
      In addition
      f(-2)=0=f(2)
      Therefore, f(x) satisfies the criteria of Roll’s theorem. Since, given function is
      \( f(x)=x^3-4x\)
      Differentiating the function w.r. to x, we get
      \( f'(x) = 3x^2-4\)
      According to Rolle's theorem, there exists at least one value \( c\in (-2,0)\) such that f′(c)=0
      So,
      \( f′(c)=0\)
      We see that
      \(3c^2-4=0 \)
      or \( c=\pm 1.15\)
      Both points are in the interval [−2,2], and, therefore, both points satisfy the conclusion.
      The figure is shown below.
   3. \( f(x)=x^2-4x+1\) over [0,4]
   4. \( f(x)=x^3-3x^2+2x+5\) over [0,2]
   5. \( f(x)=x^{2/3}\) over [0,1]
   6. \( f(x)=x^{2/3}\) over [-1,8]
   7. \( f(x)=x+\frac{1}{x}\) over [0.5,2]
   8. \( f(x)=\sin 2 x\) over [-1,1]
   9. \( f(x)=x \sqrt{x+6}\) over [-6,0]
   10. \( f(x)= \sqrt{x-1}\) over [1,3]
2. For what value of a,m, and b does the function satisfy Roll's theorem in the interval [0,2]
   \( f(x)= \begin{cases} 3 & x =0\\ -x^2+3x+a & 0 < x < 1\\ mx+b &1 \le x \le 2 \end{cases} \)

Maxima and Minima

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Maxima and minima are known as the extrema of a function. Maxima and minima are the maximum or the minimum value of a function within the given set of ranges. For the function, under the entire range, the maximum value of the function is known as the global(absolute ) maxima and the minimum value is known as the global (absolute) minima.
There are other maxima and minima of a function, which are not the absolute maxima and minima of the function and are known as local maxima and local minima

Maxima र minima फलनको छेउ-छेउको मान भनिन्छ। Maxima र minima भनेको फलनको दायराहरूबिचमा हुने अधिकतम वा न्यूनतम मान हो। फलनको सम्पूर्ण दायराहरूबिचमा हुने अधिकतम मानलाई global( absolute ) maxima भनिन्छ र सम्पूर्ण दायराहरूबिचमा हुने न्यूनतम मानलाई विश्वव्यापी global( absolute ) minima भनिन्छ।
यस बाहेक फलनक अन्य अधिकतम/न्यूनतम मानलाई local maxima/minima भनेर चिनिन्छ।

Please note that:

1. The combination of maxima and minima is extrema.
2. global maxima is he highest point on the curve within the given range and minima is the lowest point on the curve.
3. local maxima can be less than local minima and vice-versa

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How to compute Maxima/Minima

1. Step 1: Find the turning point:
   Evaluate f ′(x) and solve f ′(x)=0.
   Say p=a,b,c,... are solutions (these are turning points).
2. Step 2: Evaluate f"(x) and find f"(p).
   If f "(p) < 0 then at x= p maxima, and f(p) is maximum value
   If f "(p)> 0 then at x=p minima, and f(p) is minimum value
   If f "(p)= 0 then test fails at x=p
   
3. Repeat this for all critical points p=a,b,c,...

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Activity 1

1. Find the local minimum and maximum values of the function
   1. \(f(x) =x^3-6x^2-36x + 4\)

      Solution 👉 Click Here

      Given function is
      \( f(x) = x^3-6x^2-36x + 4\)
      Differentiating the function w.r. to x, we get
      \( f'(x) = 3x^2- 12x- 36\)
      Again, differentiating the function w.r. to x, we get
      \( f''(x) = 6x- 12\)
      For stationary point, we set
      \( f'(x) = 0\)
      or \( 3x^2- 12x- 36 = 0\)
      or \( x^2- 4x- 12 = 0\)
      or \( x = 6, x= - 2\)

      Now, the analytical table is given below.

      	x =-2	x = 6
      \( f''(x) = 6x- 12\)	f''(-2)=-24 (-ve)	f''(6)=24 (+ve)
      Result	Maximun value	Minimum value
      \( f(x) = x^3-6x^2-36x + 4\)	f(-2)=44	f(6)=-212

      
      The function f(x) has maximum value at x=-2, the maximum value is f(-2)=44.
      The function f(x) has minimum value at x=6, the minimumm value is f(6)=-212.
      
   2. \(f(x) =3x^4-4x^3-12x^2+5\)

      Solution 👉 Click Here

      Given function is
      \( f(x) = 3x^4-4x^3-12x^2+5\)
      Differentiating the function w.r. to x, we get
      \( f'(x) = 12x^3-12x^2-24x\)
      Again, differentiating the function w.r. to x, we get
      \( f''(x) = 36x^2-24x-24\)
      For stationary point, we set
      \( f'(x) = 0\)
      or \( 12x^3-12x^2-24x = 0\)
      or \( 12x(x^2-x-2) = 0\)
      or \( x = 0,x= - 1,x=2\)

      Now, the analytical table is given below.

      	x =-1	x = 0	x = 2
      \( f''(x) = 36x^2-24x-24\)	f''(-1)=36 (+ve)	f''(0)=-24 (-ve)	f''(2)=72 (+ve)
      Result	Minimum value	Maximum value	Minimum value
      \( f(x) = 3x^4-4x^3-12x^2+5\)	f(-1)=0	f(0)=5	f(2)=-27

      
      The function f(x) has minimum value at x=-1, the minimumm value is f(-1)=0
      The function f(x) has maximum value at x=0, the maximum value is f(0)=5
      The function f(x) has minimum value at x=2, the minimumm value is f(2)=-27
      
   3. \(f(x) =x +2 \sin x;0 \le x\le 2 \pi\)

      Solution 👉 Click Here

      Given function is
      \( f(x) = x +2 \sin x\)
      Differentiating the function w.r. to x, we get
      \( f'(x) = 1 +2 \cos x\)
      Again, differentiating the function w.r. to x, we get
      \( f''(x) = -2 \sin x\)
      For stationary point, we set
      \( f'(x) = 0\)
      or \( 1 +2 \cos x = 0\)
      or \(\cos x = -\frac{1}{2}\)
      or \( x = \frac{2 \pi}{3}, x= \frac{4 \pi}{3}\)

      Now, the analytical table is given below.

      	\( x = \frac{2 \pi}{3}\)	\( x = \frac{4 \pi}{3}\)
      \( f''(x) = -2 \sin x\)	\(f''(\frac{2 \pi}{3})=-\sqrt{3}\) (-ve)	\(f''(\frac{4 \pi}{3})=\sqrt{3}\) (+ve)
      Result	Maximun value	Minimum value
      \( f(x) = x +2 \sin x\)	\(f(\frac{2 \pi}{3})=3.83\)	\(f(\frac{4 \pi}{3})=2.46\)

      
      The function f(x) has maximum value at \( x = \frac{2 \pi}{3}\), the maximum value is \(f(\frac{2 \pi}{3})=3.83\)
      The function f(x) has minimum value at \( x = \frac{4 \pi}{3}\), the minimumm value is \(f(\frac{4 \pi}{3})=2.46\)
      
   4. \(f(x) =x^3-5x+3\)
   5. \(f(x) =\frac{x}{x^2+4}\)

      Solution 👉 Click Here

      Given function is
      \( f(x) =\frac{x}{x^2+4}\)
      Differentiating the function w.r. to x, we get
      \( f'(x) =\frac{4-x^2}{(x^2+4)^2}\)
      Again, differentiating the function w.r. to x, we get
      \( f''(x) = \frac{2x(x^2-12)}{(x^2+4)^3}\)
      For stationary point, we set
      \( f'(x) = 0\)
      or \( \frac{4-x^2}{(x^2+4)^2} = 0\)
      or \( x = -2, x= 2\)

      Now, the analytical table is given below.

      	\( x =-2\)	\( x =2\)
      \( f''(x) = \frac{2x(x^2-12)}{(x^2+4)^3}\)	\(f''(-2)=1/16\) (+ve)	\(f''(2)=-1/16\) (-ve)
      Result	Minimum value	Maximum value
      \( f(x) = \frac{x}{x^2+4}\)	\(f(-2)=-1/4\)	\(f(2)=1/4\)

      
      The function f(x) has maximum value at \( x = 2\), the maximum value is \(f(2)=1/4\)
      The function f(x) has minimum value at \( x =-2\), the minimumm value is \(f(-2)=-1.4\)
      
   6. \(f(x) =x+\sqrt{1-x}\)

   Solution 👉 Click Here

   Given function is
   \( f(x) =x+\sqrt{1-x}\)
   Differentiating the function w.r. to x, we get
   \( f'(x) =1-\frac{1}{2\sqrt{1-x}}\)
   Again, differentiating the function w.r. to x, we get
   \( f''(x) =-\frac{1}{4(1-x)^{3/2}}\)
   For stationary point, we set
   \( f'(x) = 0\)
   or \( 1-\frac{1}{2\sqrt{1-x}} = 0\)
   or \( x =3/4\)
   Now, the analytical table is given below.
   	\( x =3/4\)
   \( f''(x) =-\frac{1}{4(1-x)^{3/2}}\)	\(f''(3/4)=-2\) (-ve)
   Result	Maximum value
   \( f(x) = x+\sqrt{1-x}\)	\(f(3/4)=1.25\)
   
   The function f(x) has maximum value at \( x = 3/4\), the maximum value is \(f(3/4)=1.25\)
   
2. Show that \(f(x)= x^3-3x + 4\) is maximum when \( x =-1\) , minimum when \( x = 1\) , neither when \( x = 0\) .

   Solution 👉 Click Here

   Given function is
   \( f(x) = x^3-3x + 4\)
   Differentiating the function w.r. to x, we get
   \( f'(x) = 3x^2-3\)
   Again, differentiating the function w.r. to x, we get
   \( f''(x) =6x\)
   For stationary point, we set
   \( f'(x) = 0\)
   or \(3x^2-3 = 0\)
   or \( x^2 = 2\)
   or \( x = -1, x= 1\)

   Now, the analytical table is given below.

   	x =-1	x = 1
   \( f''(x) = 6x\)	f''(-1)=-6 (-ve)	f''(1)=6 (+ve)
   Result	Maximun value	Minimum value
   \( f(x) = x^3-3x + 4\)	f(-1)=6	f(1)=2

   
   The function f(x) has maximum value at x=-1, the maximum value is f(-1)=6.
   The function f(x) has minimum value at x=1, the minimumm value is f(1)=2.
   
3. Show that the maximum value of a function \( f (x) = x + \frac{100}{x}-25\) is less than its minimum value.

   Solution 👉 Click Here

   Given function is
   \( f(x) = x + \frac{100}{x}-25\)
   Differentiating the function w.r. to x, we get
   \( f'(x) = 1-\frac{100}{x^2}\)
   Again, differentiating the function w.r. to x, we get
   \( f''(x) =\frac{200}{x^3}\)
   For stationary point, we set
   \( f'(x) = 0\)
   or \(1-\frac{100}{x^2} = 0\)
   or \( x^2 = 100\)
   or \( x = -10, x= 10\)

   Now, the analytical table is given below.

   	x =-10	x = 10
   \( f''(x) = \frac{200}{x^3}\)	f''(-10)=-0.2 (-ve)	f''(10)=0.2 (+ve)
   Result	Maximun value	Minimum value
   \( f(x) = x + \frac{100}{x}-25\)	f(-10)=-45	f(10)=-5

   
   The function f(x) has maximum value at x=-10, the maximum value is f(-10)=-45.
   The function f(x) has minimum value at x=10, the minimumm value is f(10)=-5.
   
4. Show that the maximum value of a function \( f (x) = x^{5/3}-5x^{2/3}\) at x=0 is 0. Also find its minumum value at x=2.

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Activity 2

Find the maximum and minimum values of following functions

1. \(f(x)=2x^3-3x^2-12x\)
2. \( f(x) = \frac{4}{3}x^3 + 5x^2- 6x-2\)
3. \(f(x)=2+3x-x^3\)
4. \(f(x)=200+8x^3+x^4\)
5. \(f(x)=x^4-6x^2\)
6. \(f(x)=3x^5-5x^3+3\)
7. \(f(x)=(x^2-1)^3\)
8. \(f(x)=x \sqrt{x^2+1}\)
9. \(f(x)=x-3x^{1/3}\)
10. \(f(x)=\frac{1+x^2}{1-x^2}\)
11. \(f(x)=\frac{x}{(x-1)^2}\)
12. \(f(x)=\sqrt{x^2+1}-x\)

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Activity 3

1. Find the two numbers whose sum is 16 and the sum of whose squares is minimum.

   Solution 👉 Click Here

   Let \( x\) and \( y\) be two numbers such that
   \( x + y = 16\)
   If \( S\) be the sum of the squares of the two numbers, then
   \( S = x^2 + y^2\)
   or \( S = x^2 + (16- x)^2\) \( (y = 16- x)\) भएकोले
   Differentiating the function S w.r. to x, we get
   \( \frac{dS}{dx} = 4x- 32.\)
   Again, differentiating the function S' w.r. to x, we get
   \( \frac{d^2S}{dx^2} = 4\)
   For stationary point, set
   \( \frac{dS}{dx} = 0 \)
   or \( 4x- 32 = 0\)
   or \( x = 8 \)
   Since \( \frac{d^2S}{dx^2 } = 4 > 0\) , so \( S\) is minimum when \( x = 8\)
   Now, the corresponding value of \( y\) is
   \( y = 16- 8 = 8\)
   Hence the required numbers are \( 8,8\)
2. Find the dimensions of a rectangle with perimeter 100 metres so that the area of the rectangle is a maximum.
3. A rectangular box with a square base and no top is to have a volume of 108 cubic inches. Find the dimensions for the box that require the least amount of material.
4. A farmer has 24 m of fencing and wants to fence off a rectangular field that borders a straight river. He needs no fence along the river. What are the dimensions of the field that has the largest area?
5. A farmer with 80m of fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle. What is the largest possible total area of the four pens?
6. A rectangular storage container With an open top is to have a volume of \(36 m^3\). The length of its base is twice the width. Material for the base costs Rs2 per square meter. Material for the sides costs Rs16 per square meter. Find the cost of materials for the cheapest such container.
7. Find the area of the largest rectangle that can be inscribed in a semicircle of radius r.
8. A cylindrical can is to be made to hold 1 L of oil. Find the dimensions that will minimize the cost of the metal to manufacture the can.
9. A right circular cylinder is inscribed in a right circular cone so that the center lines of the cylinder and the cone coincide. The cone has 8 cm and radius 6 cm. Find the maximum volume possible for the inscribed cylinder.
10. Suppose a company’s daily profit \(f(q)=100q-0.1q^2\) is the quantity of calculators sold per day. How many calculators should the company sell to maximise profit and what is the maximum daily profit?

Distance, Velocity and Acceleration

1. Suppose that a ball is dropped from the upper observation deck of a building 450m above the ground. If the equation of the speed is \(s(t)=4.9t^2\), what is the velocity of the ball after 5 seconds. How fas is the ball travelling when it hits the ground?
2. The equation \(s(t) = −4.9 t 2 + 49 t + 15\) gives the height in meters of an object after it is thrown vertically upward from a point 15 meters above the ground at a velocity of 49 m/sec. How high above the ground will the object reach? Solution
   At highest height
   v=0
   s'=0
   t=5 second
   the heigh is s(5)
3. If a ball is thrown into the air with a velocity of 40 ft/s, its height after t seconds is given by \(y=40t-16t^2\), find the velocity when t=2
4. The position of a particle is giveen by \(f(t)=\frac{1}{1+t}\). Find the velocity and speed after 2 seconds.

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Can you write a function whose derivative does NOT exist a some point.🗨 🙋
🕵 🗩 Yes, below is a function fiven by f(x)=|x| whose derivative does NOT exist at x=0 .

Given function is
\(f(x)= |x|\)
In the simplified form, this function can be written as
\( f(x)= \begin{cases} -x &\text { for } x < 0 \\ x &\text { for } x>0 \end{cases} \)
The derivative function is given by
\( f'(x)= \begin{cases} -1 &\text { for } x < 0 \\ 1 &\text { for } x>0 \end{cases} \)
So,
At \( x=0^{-}\), the left hand limit is
\( \displaystyle \lim_{x \to 0^{-}} f'(x)= \lim_{x \to 0^{-} } -1= -1\) LHL
At \( x=0^{+}\), the right hand limit is
\( \displaystyle \lim_{x \to 0^{+}} f'(x)= \lim_{x \to 0^{+} } 1= 1\) RHL
Since, LHL≠RHL, so, the derivative of the function f(x)=|x| does NOT exist at x=0.

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Nature of points

There are 6 types of points, they are describes as below.

1. Critical Point

   A point where the derivative of a function is either zero or undefined, is called critical points.
   In the figure below, A,B,C, and D are critical points in which

   1. f'(x)=0 at A
   2. f'(x)=0 at B
   3. f'(x) is NOT defined at C (f is not differentiable at C)
   4. f'(x)=0 at D
   Critical points are the points on the graph where the function's rate of change is altered—either a change from increasing to decreasing, in concavity, or in some unpredictable fashion.
   A critical point is
   1. stationary point if the function changes from increasing/decreasign to decreasing/increasing at that point
   2. turning point if the function changes from increasing/decreasign to decreasing/increasing at that point and and is a local minimum/maximum
   3. saddle point if the function changes both increasing/decreasign and concavity at that point
   NOTE: Maxima and mimina can occurs at critical points where first derivative is undefined.

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2. Stationary Point

   A point where the derivative of a function (the gradient/slope of a graph) is zero is called a stationary point.
   In the figure below, A,B,and D are stationary points in which

   1. f'(x)=0 at A
   2. f'(x)=0 at B
   3. f'(x)=0 at D
   Note: all stationary points are critical points, but not all critical points are stationary points.

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3. Turning Point

   A point where the derivative of a function is (the gradient/slope of a graph) is zero and function has a local maximum or local minimum (extrema point), is called a turning point.There are two types of turning points: local maximum and local minimum.

   1. If f is increasing on the left interval and decreasing on the right interval, then the stationary point is a local maximum
   2. If f is decreasing on the left interval and increasing on the right interval, then the stationary point is a local minimum

   In the figure below, C is only the turning point[Please note that A and B is not turning point] in which

   1. f'(x)=0 at A, but local extrema is NOT defined at A
   2. f'(x) is NOT defined at B
   3. f'(x)=0 at C, local minima is defined at C
   Note: all turning points are stationary points, but not all stationary points are turning points.

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4. Straight line/Constant function

   Some stationary points are neither turning points nor horizontal points of inflection. For example, every point on the graph of the equation \(y = 1\) (see Figure below), or on any horizontal line, is a stationary point that is neither a turning point nor point of inflection. It is a straight line.

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5. Point of Inflection/ Plateau point

   A point where the second derivative of a function changes sign, the graph of the function will switch from concave down to concave up, or vice versa, is called an inflection point. Assuming the second derivative is continuous, it must take a value of zero at any inflection point, although not every point where the second derivative is zero is necessarily a point of inflection.

   Mathematically, a point on a function f (x) is said to be a point of inflexion if f''(x) = 0 and f'''(x)≠0.
   At this point, the concavity changes from upward to downward or vice-versa.
   Therefore the point of inflexion is the transition between concavity of the curves.

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6. Saddle point (Horizontal point of inflection)

   A saddle point or minimax point is a point on a function where the slopes (derivatives) is zero , but which is not a local extremum of the function. Thus, a saddle point is a point which is both a stationary point and a point of inflection. Since it is a point of inflection, it is not a local extremum.

   Figure below shows the graph of the function \( f(x) = x^3\) .
   The derivative of this function is \( f'(x) = 3x^2\), so the 1st derivative f'(x) =0 when x = 0 is
   \(f'(0) = 3 \times 0^2 = 0\).
   So the function \( f(x) = x^3\) has a stationary point at x = 0.
   However, this stationary point isn’t a turning point.
   Because
   The second derivative of this function \( f''(x) = 6x\), so the second derivative \( f''(x) = 0\) when x = 0 is
   \(f''(0) = 6 \times 0 = 0\).
   So the function \( f(x) = x^3\) has a inflexional point at x = 0.
   In such case, the point is called saddle point

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Find the critical points

1. \(f(x)=8x^3+81x^2-42x-8\)

   Solution 👉 Click Here

   Given function is
   \(f(x)=8x^3+81x^2-42x-8\)
   Differentiating w. r. to x, we get
   \(f'(x)=24x^2+162x-42\)
   Solving the derivative for \(f'(x)=0\), we get
   \(f'(x)=0\)
   or\(24x^2+162x-42=0\)
   or\(x=0.25,x=-7\)
   So, the critical points are \(x=0.25,x=-7\).
2. \(f(x)=1+80x^3+5x^4-2x^5\)

   Solution 👉 Click Here

   Given function is
   \(f(x)=1+80x^3+5x^4-2x^5\)
   Differentiating w. r. to x, we get
   \(f'(x)=240x^2+20x^3-10x^4\)
   Solving the derivative for \(f'(x)=0\), we get
   \(f'(x)=0\)
   or\(240x^2+20x^3-10x^4=0\)
   or\(x^2=0, 24+2x-x^2=0\)
   or\(x=-4,x=0,x=6\)
   So, x=-4,x=0,x=6 are critical points.
   
3. \(f(x)=2x^3-7x^2-3x-2\)
4. \(f(x)=x^6-2x^5+8x^4\)
5. \(f(x)=4x^3-3x^2+9x+12\)
6. \(f(x)=\frac{x+4}{2x^2+x+8}\)

   Solution 👉 Click Here

   Given function is
   \(f(x)=\frac{x+4}{2x^2+x+8}\)
   Differentiating w. r. to x, we get
   \(f'(x)=\frac{-2(x^2+8x-2)}{(2x^2+x+8)^2}\)
   Solving the derivative for numirator=0, we get
   numirator=0
   or\(-2(x^2+8x-2)=0\)
   or\(x^2+8x-2\)
   or\(x=0.24,x=-8.24\)
   So, the critical points are \(x=0.24,x=-8.24\). Next, solving the derivative for denomirator=0, we get
   denomirator=0
   or\((2x^2+x+8)^2=0\)
   or\(2x^2+x+8=0\)
   orno real solution
   Hence, the critical points are \(x=0.24,x=-8.24\).
7. \(f(x)=\frac{1-x}{x^2+2x-15}\)

   Solution 👉 Click Here

   Given function is
   \(f(x)=\frac{1-x}{x^2+2x-15}\)
   Differentiating w. r. to x, we get
   \(f'(x)=\frac{x^2-2x+13}{(x^2+2x-15)^2}\)
   Solving the derivative for numirator=0, we get
   numirator=0
   or\(x^2-2x+13=0\)
   orno real solution
   Next, solving the derivative for denomirator=0, we get
   denomirator=0
   or\((x^2+2x-15)^2=0\)
   or\(x^2+2x-15=0\)
   or\(x=-5,x=3\)
   As we know, a critical point (derivative zero, DNE: derivative not exist) must exist as a point of original function. For this problem we found two values x=-5,x=3 where the derivative doesn’t exist, and the function also doesn’t exist, so neither of these are critical points.
   So, \(f(x)=\frac{1-x}{x^2+2x-15}\) have no critical points.
8. \(f(x)=\sqrt[5]{x^2-6x}\)

   Solution 👉 Click Here

   Given function is
   \(f(x)=\sqrt[5]{x^2-6x}\)
   Differentiating w. r. to x, we get
   \(f'(x)=\frac{2x-6}{5(x^2-6x)^{4/5}}\)
   Solving the derivative for numirator=0, we get
   numirator=0
   or\(2x-6=0\)
   or\(x=3\)
   Next, solving the derivative for denomirator=0, we get
   denomirator=0
   or\(5(x^2-6x)^{4/5}=0\)
   or\(x^2-6x=0\)
   or\(x=0,x=6\)
   As we know, a critical point (derivative zero, DNE: derivative not exist) must exist as a point of original function. For this problem we found two values x=0,x=3,x=6 exist in the function, so these x=0,x=3,x=6 are critical points.
   

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Find the inflexional point

1. Determine the concavity of \(f(x) = x^ 3 − 6 x^ 2 −12 x + 2\) and identify any points of inflection.

   Solution 👉 Click Here

   Given function is
   \(f(x) = x^ 3 − 6 x^ 2 −12 x + 2\)
   Differentiating w. r. to x, we get
   \(f'(x) = 3x^ 2 − 12 x −12\)
   Now, the second derivative is
   \(f''(x)=6x-12\)
   Solving the derivative for \(f''(x)=0\), we get
   \(f''(x)=0\)
   or\(6x-12=0\)
   or\(x=2\)
   The inflexional point is x=2.

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Find the concavity

1. Determine the concavity of \(f(x) = \sin x + \cos x\) on [0,2π] and identify any points of inflection

   Solution 👉 Click Here

   Given function is
   \(f(x) = \sin x + \cos x\)
   Differentiating w. r. to x, we get
   \(f'(x) = \cos x - \sin x\)
   Now, the second derivative is
   \(f''(x) = - \sin x - \cos x\)
   Solving the derivative for \(f''(x)=0\), we get
   \(f''(x)=0\)
   or\(- \sin x - \cos x=0\)
   or\(x=135^0, 315^0\)
   The inflexional point is \(x=\frac{3 \pi}{4}\) and \(x=\frac{7 \pi}{4}\) . Now

   interval	\((0,\frac{3 \pi}{4})\)	\((\frac{3 \pi}{4},\frac{7 \pi}{4})\)	\((\frac{7 \pi}{4},2\pi)\)
   \(f''(x)=- \sin x - \cos x\)	(-ve)	(+ve)	(-ve)
   result	down	up	down

   So, the curve is concave down during \((0,\frac{3 \pi}{4})\), concave up during \((\frac{3 \pi}{4},\frac{7 \pi}{4})\), and concave down during \((\frac{7 \pi}{4},2\pi)\).
2. Discuss the curve/function with respect to cancavity, point of inflexion, and local maxima, minima.
   1. \(y=x^4-4x^3\)
   2. \(y=x^{2/3}(6-x)^{1/3}\)
   3. \( f(x) = \frac{1}{6}x^4-2x^3+11x^2-18x\)
   4. \( f(x) = \frac{1}{12}x^4-2x^2\)
   5. \( f(x) = x^3-3x^2+1\)
   6. \( f(x) = \frac{2}{3}x^3-\frac{5}{2}x^2+2x\)

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Question for Practice

1. Find the stationary points of the function \( f(x) = 2x^3-3x^2-36x\).
2. Find the approximate values, to two decimal places, of the stationary points of the function \( f(x) = x^3-x^2-2x\).
3. Find the (a) intervals of increasing/decreasing (b) intervals of concave upward/downward (c) point of inflection
   1. \(f(x)=2x^3-3x^2-12x\)
   2. \( f(x) = \frac{4}{3}x^3 + 5x^2- 6x-2\)
   3. \(f(x)=2+3x-x^3\)
   4. \( f(x) = 3x^4 -2x^3- 9x^2 + 7\)
   5. \(f(x)=x^4-6x^2\)
   6. \(f(x)=200+8x^3+x^4\)
   7. \(f(x)=3x^5-5x^3+3\)
   8. \(f(x)=(x^2-1)^3\)
   9. \(f(x)=x \sqrt{x^2+1}\)
   10. \(f(x)=x-3x^{1/3}\)
4. Find the (a) vertical and horizental asymptotes (b) intervals of increasing/decreasing (c) intervals of concave upward/downward (d)point of inflection of following functions
   1. \(f(x)=\frac{1+x^2}{1-x^2}\)
   2. \(f(x)=\frac{x}{(x-1)^2}\)
   3. \(f(x)=\sqrt{x^2+1}-x\)

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Find the local Extreme

1. Find any turning points of the function \(f(x)=x^3-10x^2+25x+4\) and specify whether each turning point is a global maximum or minimum, or a local maximum or minimum.
   

   Solution 👉 Click Here

   Given function is
   \(f(x)=x^3-10x^2+25x+4\)
   Differentiating w. r. to x, we get
   \(f'(x)=3x^2-20x+25\)
   Solving the derivative for \(f'(x)=0\), we get
   \(f'(x)=0\)
   or\(3x^2-20x+25=0\)
   or\(x=5/3,x=5\)
   So, x=5/3,x=5 are critical points.
   Now, the second derivative is
   \(f''(x)=6x-20\)
   Now

   point	derivative	value	result
   at x=5/3	f''(x)=6x-20	-10 (-ve)	maxima
   at x=5	f''(x)=6x-20	10 (+ve)	minima

   So, x=5/3 is local maxima and x=5 is local minima.

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Question for Exam

1. Suppose you are given formula for a function f
   1. How do you determine where f is increasing/decreasing
   2. How do you determine where f is concave upward/downward
   3. How do you locate inflexional points
2. Find the critical numbers of \(f(x)=x^4(x-1)^3\).What does the second derivative tells you about the behavior of f at these points. What does the first derivative test tell you?