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 theore \[f\left( x \right) = \sqrt{25 - x^2}\] on [−3, 4] ?

Answer 1

We have,

\[f\left( x \right) = \sqrt{25 - x^2}\]

Here, \[f\left( x \right)\] will exist,
if  \[25 - x^2 \geq 0\]

\[ \Rightarrow x^2 \leq 25\]

\[ \Rightarrow - 5 \leq x \leq 5\]

Since for each \[x \in \left[ - 3, 4 \right]\] , the function \[f\left( x \right)\] attains a unique definite value.

So,\[f\left( x \right)\] is continuous on \[\left[ - 3, 4 \right]\]

Also, \[f'\left( x \right) = \frac{1}{2\sqrt{25 - x^2}}\left( - 2x \right) = \frac{- x}{\sqrt{25 - x^2}}\] exists for all \[x \in \left( - 3, 4 \right)\]

 so ,\[f\left( x \right)\] is differentiable on \[\left( - 3, 4 \right)\] .

Thus, both the conditions of lagrange's theorem are satisfied.
Consequently, there exists some 

\[c \in \left( - 3, 4 \right)\]  such that

\[f'\left( c \right) = \frac{f\left( 4 \right) - f\left( - 3 \right)}{4 + 3} = \frac{f\left( 4 \right) - f\left( - 3 \right)}{7}\]

Now,

\[f\left( x \right) = \sqrt{25 - x^2}\]

\[f'\left( x \right) = \frac{- x}{\sqrt{25 - x^2}}\] ,

\[f\left( - 3 \right) = 4\] ,

\[f\left( 4 \right) = 3\]

∴ \[f'\left( x \right) = \frac{f\left( 4 \right) - f\left( - 3 \right)}{4 + 3}\]

\[\Rightarrow \frac{- x}{\sqrt{25 - x^2}} = \frac{3 - 4}{7}\]

\[ \Rightarrow 49 x^2 = 25 - x^2 \]

\[ \Rightarrow x = \pm \frac{1}{\sqrt{2}}\]

Thus, \[c = \pm \frac{1}{\sqrt{2}} \in \left( - 3, 4 \right)\] such that

\[f'\left( c \right) = \frac{f\left( 4 \right) - f\left( - 3 \right)}{4 - \left( - 3 \right)}\] .

Hence, Lagrange's theorem is verified.