Question

Given the function $f\left( x \right) = \sin x$, how do you determine whether $f$ satisfies the hypotheses of the Mean Value Theorem on the interval $\left[ {0,\pi } \right]$ and find the $c$?

Answer 1

Hint: In this of questions first we should check whether the given function is continuous on the given closed interval and differentiable on the given open interval, and then if the function satisfies the Mean Value Theorem then by using the definition of the theorem we will find the value of $c$.

Complete step by step solution:
Given function is $f\left( x \right) = \sin x$, and the interval in which satisfies Mean Value Theorem is $\left[ {0,\pi } \right]$,
Mean value theorem is defined as, If a function f is continuous on $\left[ {a,b} \right]$ and differentiable on $\left( {a,b} \right)$ then there exists $c$ in $\left( {a,b} \right)$ such that $f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}$,
Now in the given function $f\left( x \right) = \sin x$ we can see that the function is continuous and differentiable at all points, therefore the given function is continuous on the closed interval $\left[ {0,\pi } \right]$ and differentiable on the open interval $\left( {0,\pi } \right)$, which means we can conclude that there exists at least one point $c$in the interval $\left( {0,\pi } \right)$ such that $f'\left( c \right)$ is equal to the functions average rate of change over $\left[ {0,\pi } \right]$.
Now from the given data,
$ \Rightarrow f\left( x \right) = \sin x$,
Now differentiating on both sides we get,
$ \Rightarrow f'\left( x \right) = \dfrac{d}{{dx}}\sin x$,
Now simplifying we get,
$ \Rightarrow f'\left( x \right) = \cos x - - - - (1)$,
Now the given interval is $\left[ {0,\pi } \right]$, so according to the theorem, $a = 0$ and $b = \pi $, so applying the Mean Value Theorem we get,
$ \Rightarrow f'\left( c \right) = \dfrac{{\sin \pi - \sin 0}}{{\pi - 0}}$,
Now simplifying we get,
$ \Rightarrow f'\left( c \right) = \dfrac{{0 - 0}}{{\pi - 0}}$,
Further simplifying we get,
$ \Rightarrow f'\left( c \right) = \dfrac{0}{\pi } = 0 - - - - - (2)$,
Now comparing equations (1) and (2) we get,
$ \Rightarrow \cos c = 0$,
Now from the trigonometric ratios we get,
$ \Rightarrow \cos c = \cos \dfrac{\pi }{2}$,
Now from then above we can conclude that,
$ \Rightarrow c = \dfrac{\pi }{2}$,

$\therefore $ It is proved the function $f\left( x \right) = \sin x$, how do you determine whether $f$ satisfies the hypotheses of the Mean Value Theorem on the interval $\left[ {0,\pi } \right]$, and the value of $c$ will be equal to $\dfrac{\pi }{2}$.

Note:
We should know that the Mean Value Theorem doesn’t tell us what $c$ is. It only tells us that there is at least one number $c$ that will satisfy the conclusion of the theorem.