###### Increasing Function

A function \( y = f (x)\) is said to be increasing in an open interval \( (a, b)\) i.e. between the points \( x = a\) and \( x = b\) if

\( \frac{dy}{dx} \) or \( f ' (x)\) is positive

for all values of \( x\) in that interval.

In other words, f(x) is said to be increasing if

\( x_1 < x_2 \Rightarrow f (x_1) < f (x_2)\)

or equivalently

\( x_1 > x_2 \Rightarrow f (x_1) > f (x_2)\)

for all real numbers \( x_1\) and \( x_2\) in \( (a, b)\) .

An increasing function can be shown graphically as in the figure below.

The graph of such a function is a curve which goes on rising with the rise in the value of \( x\).

###### Decreasing Function

A function \( y = f (x)\) is said to be decreasing in an open interval \( (a, b)\) i.e. between the points \( x = a\) and \( x = b\) if

\( \frac{dy}{dx} \) or \( f ' (x)\) is negative

for all values of \( x\) in that interval.

In other words, f(x) is said to be increasing if

\( x_1 < x_2 \Rightarrow f (x_1) > f (x_2)\)

or equivalently

\( x_1 > x_2 \Rightarrow f (x_1) < f (x_2)\)

for all real numbers \( x_1\) and \( x_2\) in \( (a, b)\) .

A decreasing function can be shown graphically as in the figure below.

The graph of such a function is a curve which goes on falling with the raise in the value of \( x\).

###### Example 1

Show that the function \( f(x) = 2x^3 - 3x^2 - 36x \) is increasing on each of the intervals (-∞,-2) and (3,∞), and decreasing on the interval (-2, 3).

Solution

Find the derivative. The derivative is

\( f'(x) = 6x^2 - 6x - 36\) (1)

Factorizing (i) gives

\( f'(x) = 6(x^2 - x - 6) = 6(x + 2)(x - 3)\)

(-∞,-2) | (-2,3) | (3,∞) | |

(x+2) | - | + | + |

(x-3) | - | - | + |

f'(x) | + | - | + |

curve | increasing | decreasing | increasing |

- When x is less than -2, the values of x + 2 and x - 3 are both negative, and hence the value of \( f'(x) = 6(x + 2)(x - 3) \) is positive.

Therefore, by the increasing/decreasing criterion, the function f is increasing on the interval (-∞,-2). - When x is in the interval (-2, 3), the value of x + 2 is positive and the value of x - 3 is negative, and hence the value of \( f'(x) = 6(x + 2)(x - 3)\) is negative.

Therefore, by the increasing/decreasing criterion, the function f is decreasing on the interval (-2, 3). - When x is greater than 3, the values of x + 2 and x - 3 are both positive, and hence the value of \( f'(x) = 6(x + 2)(x - 3)\) is also positive.

Therefore, by the increasing/decreasing criterion, the function f is increasing on the interval (3,∞).

#### Examples : Activity 1

- Examine whether the function
- \(f(x)=15x^2-14x+1\) is increasing or decreasing at \(x=\frac{2}{5}\) and \(x=\frac{5}{2}\).
- \(f(x)= 2x^3-24x+15 \) is increasing or decreasing at \(x= 3 \) and \(x= \frac{3}{2} \).
- \(f(x)= 2x^2-4x+3 \) is increasing or decreasing at (1,4).
- \(f(x)= 16x-\frac{4}{3}x^3 \) is increasing or decreasing at \((- \infty ,-2)\).

- Show that the function given by
- \( f(x) =- x ^3 + 6x^ 2 -13x+20, x \in R\) is decreasing on R.
- \( f(x) = x ^3 – 3x^ 2 + 4x, x \in R\) is increasing on R.
- \(f (x) = x^ 3 – 3x^ 2 + 3x – 100\) is increasing in R
- \( f(x) = 4x-\frac{9}{x}+6 \) is increasing on R except x=0.

- Find the intervals in which the following functions are strictly increasing or decreasing:
- \(f(x) = x^ 2 – 4x + 6\)
- \(f(x) =x ^2 + 2x – 5 \)
- \(f(x) =3x ^2 -6x + 5 \)
- \(f(x) =10 – 6x – 2x^ 2\)
- \(f(x) = 2x^ 2 – 3x\)
- \(f(x) =6 – 9x – x ^2 \)
- \(f(x) = x^3 − 3x^2 + 4x + 3\)
- \(f(x) = (x + 1)^3 (x – 3) \)
- \(f(x) =–2x ^3 – 9x^ 2 – 12x + 1 \)
- \( f(x) = \frac{2}{3}x^3 - 8x^2 + 30x - 36\)
- \(f(x) =3x^4-4x^3-12x^2+5\)
- \(f (x) = 4x ^3 – 6x^ 2 – 72x + 30\)
- \(f(x) =5x ^3 – 135x + 22 \)
- \(f(x) =6+12x+3x^2-2x^3 \)
- \(f(x) = 2x ^3 – 3x ^2 – 36x + 7\)
- \(f (x) = \sin 3x, x \in [0,\pi/2]\)
- \(f(x) = \sin x + \cos x, 0 \le x \le 2 \pi \)

- Prove that the function given by \(f(x) = \cos x\) is (a) decreasing in (0, π) (b) increasing in (π, 2π), and (c) neither increasing nor decreasing in (0, 2π).

## No comments:

## Post a Comment