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

\(f '(c) = \frac{f(b)-f(a)}{b-a}\)

#### 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.

#### Proof of Mean value theorem

Statementif 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

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

- Which of the following function satisfy the hypothesis of Mean Value theorem, give reason.
- \( f(x)=x^2\) over [0,2]
- \( f(x)=x^2-5x+1\) over [0,3]
- \( f(x)=x^2-2x-8\) over [-1,3]
- \( f(x)=x^3+3x^2-24x\) over [1,4]
- \( f(x)=4x^3-8x^2+7x-2\) over [2,5]
- \( f(x)=x^3-8x-5\) over [1,4]
- \( f(x)=2^x\) over [0,3]
- \( f(x)=x^{2/3}\) over [-1,8]
- \( f(x)=x^{4/5}\) over [0,1]
- \( f(x)=\frac{x+2}{x}\) over [1,2]
- \( f(x)=\frac{2}{x}\) over [-1,1]
- \( f(x)=5-\frac{4}{x}\) over [1,4]
- \( f(x)=x \sqrt{1-x}\) over [0,1]
- \( f(x)=2 \sin x+3 \cos x\) over \([0,2 \pi]\)
- \( f(x)= \begin{cases} \frac{\sin x}{x} & \pi \le x < 0 \\ 0 & x=0 \end{cases} \)

- For what value of a,m, and b does the function satisfy Mean value theorem in the interval [0,2]

\( f(x)= \begin{cases} 3 & x =0\\ -x^2+3x+a & 0 < x < 1\\ mx+b &1 \le x \le 2 \end{cases} \)

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