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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] ?
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Solution
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 \[\left[ - 3, 4 \right]\]
Consequently, there exists some
\[\Rightarrow \frac{- x}{\sqrt{25 - x^2}} = \frac{3 - 4}{7}\]
\[ \Rightarrow 49 x^2 = 25 - x^2 \]
\[ \Rightarrow x = \pm \frac{1}{\sqrt{2}}\]
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.
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