Name: Example of Mean Value Theorem
Uploaded: 2019-07-19
Duration: 2 min 17 s
Description: You've already learned about the Mean Value Theorem. Let’s explore more about the theorem with some examples in this video. Watch and learn!

Question

Verify Lagrange's mean value theorem for the following function on the indicated intervals. Find a point $' c'$ in the indicated interval as stated by the Lagrange's mean value theorem:

$f (x) = \sqrt{25 - x^{2}}$ on $[- 3, 4]$

Accepted Answer

Given

$f (x) = \sqrt{25 - x^{2}}$

Here, $f (x)$ will exist,

If $25 - x^{2} \geq 0 \Rightarrow x^{2} \leq 25$

$\Rightarrow - 5 \leq x \leq 5$

Hence, $f (x)$ is defined in the given interval $[- 3, 4]$ .

So, $f (x)$ is continuous on $[- 3, 4]$ .

Use chain rule of differentiation

$f^{'} (x) = \frac{d}{d x} (f (x)) = \frac{1}{2 \sqrt{25 - x^{2}}} (- 2 x) = \frac{- x}{\sqrt{25 - x^{2}}}$ exists for all $x \in (- 3, 4)$

So, $f (x)$ is differentiable on $(- 3, 4)$

Thus, both the conditions of lagrange's theorem are satisfied.

Consequently, there exists some $c \in (- 3, 4)$ such that $f^{'} (c) = \frac{f (4) - f (- 3)}{4 + 3} = \frac{f (4) - f (- 3)}{7}$

Now $f (x) = \sqrt{25 - x^{2}}$

$f^{'} (x) = \frac{- x}{\sqrt{25 - x^{2}}}, f (- 3) = 4, f (4) = 3$

$∴ f^{'} (c) = \frac{f (4) - f (- 3)}{7}$

$\Rightarrow \frac{- c}{\sqrt{25 - c^{2}}} = \frac{3 - 4}{7}$

$\Rightarrow 49 c^{2} = 25 - c^{2}$

$\Rightarrow c = \pm \frac{1}{\sqrt{2}}$

Thus, $c = \pm \frac{1}{\sqrt{2}} \in (- 3, 4)$ Such that $f^{'} (c) = \frac{f (4) - f (- 3)}{4 - (- 3)}$

Hence, Lagrange's theorem is verified.

R D Sharma Solutions for Chapter: Mean Value Theorems, Exercise 1: EXERCISE

R D Sharma Mathematics Solutions for Exercise - R D Sharma Solutions for Chapter: Mean Value Theorems, Exercise 1: EXERCISE

Questions from R D Sharma Solutions for Chapter: Mean Value Theorems, Exercise 1: EXERCISE with Hints & Solutions

Learn and Prepare for any exam you want !