Question

How do you determine if Rolle's theorem can be applied to $f\left( x \right) = \sin 2x$ on the interval $\left[ {0,\dfrac{\pi }{2}} \right]$ and if so, how do you find all the values of $c$ in the interval for which $f'\left( c \right) = 0$?

Answer 1

Hint: In order to solve for Rolle's theorem, first we should know what Rolle's theorem is, then according to Rolle's theorem, prove the first three necessary rules. Equate the derivative of the given function with zero and get the value for $c$ in the given interval $\left( {a,b} \right)$.

Formula used:
1) $f'\left( {\sin x} \right) = \cos x$
2) $\sin 0 = 0$
3) $\sin \pi = 0$
4) $\cos \dfrac{\pi }{2} = 0$

Complete step by step solution:
We are given with the function: $f\left( x \right) = \sin 2x$.
Since, we know that, the Rolle's theorem says that if:
1) $y = f\left( x \right)$ is a continuous function in a set $\left[ {a,b} \right]$;
2) $y = f\left( x \right)$ is a derivable function in a set $\left( {a,b} \right)$;
3) $f\left( a \right) = f\left( b \right)$; Then, there exists a $c \in \left( {a,b} \right)$as if $f'\left( c \right) = 0$ exists.
So, for solving our function with Rolle's theorem, we need to check all the three steps and are as follows:
1) $y = f\left( x \right) = \sin 2x$, since we know that the function sine is continuous everywhere, so according to that $f\left( x \right) = \sin 2x$ is also continuous in the interval $\left[ {0,\dfrac{\pi }{2}} \right]$. That implies the first point proved.
2) Differentiating $f\left( x \right) = \sin 2x$ and we get $f'\left( x \right) = 2\cos 2x$, which is also continuous everywhere, so our function is also differentiable everywhere. That gives our function is also differentiable at $\left( {0,\dfrac{\pi }{2}} \right)$.
3) Substituting $0$ in $f\left( x \right) = \sin 2x$ and we get: $f\left( 0 \right) = \sin 2\left( 0 \right) = 0$, Then substituting $\dfrac{\pi }{2}$ in $f\left( x \right) = \sin 2x$ and we get: $f\left( {\dfrac{\pi }{2}} \right) = \sin 2.\dfrac{\pi }{2} = \sin \pi = 0$. And we obtained that $f\left( a \right) = f\left( b \right) \\
\Rightarrow f\left( 0 \right) = f\left( {\dfrac{\pi }{2}} \right) \\
 $
And, this way our Rolle's theorem is verified, so there exists a $c \in \left( {a,b} \right)$as if $f'\left( c \right) = 0$exists.
Now, derivating $f\left( x \right) = \sin 2x$ which can be written as $f\left( c \right) = \sin 2c$, with respect to $c$, and we get:
$f'\left( c \right) = 2\cos 2c$
Comparing $f'\left( c \right)$ with zero, $f'\left( c \right) = 0$:
$
  f'\left( c \right) = 0 \\
  2\cos 2c = 0 \\
  \cos 2c = 0 \\
 $
Since, we know that $\cos \dfrac{\pi }{2} = 0$, so substituting it in the above equation, we get:
$
  \cos 2c = 0 \\
  \cos 2c = \cos \dfrac{\pi }{2} \\
\Rightarrow 2c = \dfrac{\pi }{2} \\
\Rightarrow c = \dfrac{\pi }{4} \\
 $
Since, we can see that $c = \dfrac{\pi }{4}$lies in the interval $c \in \left( {0,\dfrac{\pi }{2}} \right)$.
Therefore, Rolle's theorem can be applied to the function to $f\left( x \right) = \sin 2x$ on the interval $\left[ {0,\dfrac{\pi }{2}} \right]$ and value of $c$ for which $f'\left( c \right) = 0$ is $c = \dfrac{\pi }{4}$ in the interval $\left( {0,\dfrac{\pi }{2}} \right)$.

Note:
1) It’s important to check for the Rolle's theorem rules step by step, to solve this kind of function.
2) The square bracket represents the closed interval whereas the round bracket represents the open intervals.