Question

The value of $c$ from the Lagrange’s mean value theorem for which $f(x) = \sqrt {25 - {x^2}}$ in $[1,5]$ is
A) $5$
B) $1$
C) $\sqrt {15}$
D) None of these

Answer 1

First of all we have to know what Lagrange's mean value theorem or LMVT. So, Lagrange’s mean value theorem or LMVT states that if a function $f(x)$ is continuous on a closed interval $[a,b]$ and differentiable on the open interval $\left( {a,b} \right)$, then there is at least one point $x = c$ on this interval, such that ,
$f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{\left( {b - a} \right)}}$
We are going to use this formula to find the value of $c$ for the given function.

Complete step by step answer:
Given, $f(x) = \sqrt {25 - {x^2}}$ in the interval $[1,5]$.
So, in the f=given interval, we can clearly see that $f(x)$ is continuous.
So, our first condition for Lagrange’s mean value theorem stands true.
Now, on differentiating $f(x)$ with respect to $x$, we get,
$f'\left( x \right) = \dfrac{d}{{dx}}\left( {\sqrt {25 - {x^2}} } \right)$
$\Rightarrow f'\left( x \right) = \dfrac{1}{{2\sqrt {25 - {x^2}} }}\dfrac{d}{{dx}}\left( { - {x^2}} \right)$
[Using chain rule]
$\Rightarrow f'\left( x \right) = \dfrac{1}{{2\sqrt {25 - {x^2}} }}\left( { - 2x} \right)$
Now, by simplifying the equation, we get,
$\Rightarrow f'\left( x \right) = \dfrac{{ - x}}{{\sqrt {25 - {x^2}} }}$
So, we can clearly observe that, $f'\left( x \right)$ is differentiable in $\left( {1,5} \right)$.
So, the second condition for Lagrange’s mean value theorem is also true.
Therefore, there must be a $c \in \left[ {1,5} \right]$ such that,
$f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{\left( {b - a} \right)}}$
Here, $b = 5$ and $a = 1$.
So, $f'\left( c \right) = \dfrac{{f\left( 5 \right) - f\left( 1 \right)}}{{\left( {5 - 1} \right)}}$
Now, opening the functions, we get,
\Rightarrow f'\left( c \right) = \dfrac{{\left\\{ {\sqrt {25 - {{\left( 5 \right)}^2}} } \right\\} - \left\\{ {\sqrt {25 - {{\left( 1 \right)}^2}} } \right\\}}}{{\left( 4 \right)}}
\Rightarrow f'\left( c \right) = \dfrac{{\left\\{ {\sqrt {25 - 25} } \right\\} - \left\\{ {\sqrt {25 - 1} } \right\\}}}{{\left( 4 \right)}}
Simplifying the equation, we get,
\Rightarrow f'\left( c \right) = \dfrac{{0 - \left\\{ {\sqrt {24} } \right\\}}}{{\left( 4 \right)}}
$\Rightarrow f'\left( c \right) = \dfrac{{ - 2\sqrt 6 }}{4}$
$\Rightarrow f'\left( c \right) = \dfrac{{ - \sqrt 6 }}{2}$
But, we know, $f'\left( x \right) = \dfrac{{ - x}}{{\sqrt {25 - {x^2}} }}$.
Therefore, $f'\left( c \right) = \dfrac{{ - c}}{{\sqrt {25 - {c^2}} }}$
So, $\dfrac{{ - c}}{{\sqrt {25 - {c^2}} }} = \dfrac{{ - \sqrt 6 }}{2}$
Now, squaring both sides, we get,
$\Rightarrow \dfrac{{{c^2}}}{{25 - {c^2}}} = \dfrac{6}{4}$
Cross multiplying the above equation, we get,
$\Rightarrow 4{c^2} = 6\left( {25 - {c^2}} \right)$
$\Rightarrow 4{c^2} = 150 - 6{c^2}$
Adding $6{c^2}$ on both sides, we get,
$\Rightarrow 10{c^2} = 150$
$\Rightarrow {c^2} = 15$
Now, square rooting both sides, we get,
$\Rightarrow c = \pm \sqrt {15}$
Therefore, $c = \sqrt {15} \in \left[ {1,5} \right]$. Hence, the correct option is (C).

Note:
There are many important applications of Lagrange’s mean value theorem. It plays an important role in the proof of Fundamental Theorem of Calculus. The generalisation of the Lagrange’s mean value theorem can be used to prove L’Hopital’s Rule. Mean value theorem is important in other areas of mathematics other than calculus and analysis, like applied mathematics and even number theory.