Verify Rolle's theorem for each of the following functions on the indicated intervals

Solution:

(i) The given function is $f(x)=\cos 2\left(x-\frac{\pi}{4}\right)=\cos \left(2 x-\frac{\pi}{2}\right)=\sin 2 x$.

Since $\sin 2 x$ is everywhere continuous and differentiable.

Therefore, $\sin 2 x$ is continuous on $\left[0, \frac{\pi}{2}\right]$ and differentiable on $\left(0, \frac{\pi}{2}\right)$.

Also,

$f\left(\frac{\pi}{2}\right)=f(0)=0$

Thus, $f(x)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c \in\left(0, \frac{\pi}{2}\right)$ such that $f^{\prime}(c)=0$.

We have

$f(x)=\sin 2 x$

$\Rightarrow f^{\prime}(x)=2 \cos 2 x$

$\therefore f^{\prime}(x)=0$

$\Rightarrow 2 \cos 2 x=0$

$\Rightarrow \cos 2 x=0$

 

$\Rightarrow x=\frac{\pi}{4}$

Thus, $c=\frac{\pi}{4} \in\left(0, \frac{\pi}{2}\right)$ such that $f^{\prime}(c)=0$.

Hence, Rolle's theorem is verified.

(ii) The given function is $f(x)=\sin 2 x$.

Since $\sin 2 x$ is everywhere continuous and differentiable.

Therefore, $\sin 2 x$ is continuous on $\left[0, \frac{\pi}{2}\right]$ and differentiable on $\left(0, \frac{\pi}{2}\right)$.

Also,

$f\left(\frac{\pi}{2}\right)=f(0)=0$

Thus, $f(x)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c \in\left(0, \frac{\pi}{2}\right)$ such that $f^{\prime}(c)=0$.

We have

$f(x)=\cos 2 x$

$\Rightarrow f^{\prime}(x)=-2 \sin 2 x$

$\therefore f^{\prime}(x)=0$

$\Rightarrow-2 \sin 2 x=0$

$\Rightarrow \sin 2 x=0$

$\Rightarrow \sin 2 x=0$

 

$\Rightarrow \mathrm{x}=0$

Since $c=0 \in\left(\frac{-\pi}{4}, \frac{\pi}{4}\right)$ such that $f^{\prime}(c)=0$.

​Hence, Rolle's theorem is verified.

(iv)

The given function is $f(x)=e^{x} \sin x$.

Since $\sin x \& e^{x}$ are everywhere continuous and differentiable.

Therefore, being a product of these two, $f(x)$ is continuous on $[0, \pi]$ and differentiable on $(0, \pi)$.

Also,

$f(\pi)=f(0)=0$

Thus, $f(x)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c \in(0, \pi)$ such that $f^{\prime}(c)=0$.

We have

$f(x)=e^{x} \sin x$

 

$\Rightarrow f^{\prime}(x)=e^{x}(\sin x+\cos x)$

$\therefore f^{\prime}(x)=0$

$\Rightarrow e^{x}(\sin x+\cos x)=0$

$\Rightarrow \sin x+\cos x=0$

$\Rightarrow \tan x=-1$

 

$\Rightarrow \mathrm{x}=\pi-\frac{\pi}{4}=\frac{3 \pi}{4}$

Since $c=\frac{3 \pi}{4} \in(0, \pi)$ such that $f^{\prime}(c)=0$.

​Hence, Rolle's theorem is verified.

(v)

The given function is $f(x)=e^{x} \cos x$

Since $\cos x \& e^{x}$ are everywhere continuous and differentiable, $f(x)$ being a product of these two is continuous on $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$ and differentiable on $\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$.

Also, 

$f\left(\frac{-\pi}{2}\right)=f\left(\frac{\pi}{2}\right)=0$

Thus, $f(x)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$ such that $f^{\prime}(c)=0$.

We have

$f(x)=e^{x} \cos x$

 

$\Rightarrow f^{\prime}(x)=e^{x}(\cos x-\sin x)$

$\therefore f^{\prime}(x)=0$

$\Rightarrow e^{x}(\cos x-\sin x)=0$

$\Rightarrow \sin x-\cos x=0$

$\Rightarrow \tan x=1$

 

$\Rightarrow \mathrm{x}=\frac{\pi}{4}$

Since $c=\frac{\pi}{4} \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$ such that $f^{\prime}(c)=0$

Hence, Rolle's theorem is verified.

(vi)

The given function is $f(x)=\cos 2 x$.

Since $\cos 2 x$ is everywhere continuous and differentiable

Therefore, $f(x)$ is continuous on $[0, \pi]$ and differentiable on $(0, \pi)$.

Also,

$f(\pi)=f(0)=1$

Thus, $f(x)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c \in(0, \pi)$ such that $f^{\prime}(c)=0$.

We have

$f(x)=\cos 2 x$

 

$\Rightarrow f^{\prime}(x)=-2 \sin 2 x$

$\therefore f^{\prime}(x)=0$

$\Rightarrow-2 \sin 2 x=0$

$\Rightarrow \sin 2 x=0$

$\Rightarrow 2 x=\pi$

 

$\Rightarrow \mathrm{x}=\frac{\pi}{2}$

Thus, $c=\frac{\pi}{2} \in(0, \pi)$ such that $f^{\prime}(c)=0$.

​Hence, Rolle's theorem is verified.

(vii)

The given function is $f(x)=\frac{\sin x}{e^{x}}$.

Since $\cos x$ and $e^{x}$ are everywhere continuous and differentiable, being the quotient of these two, $f(x)$ is continuous on $[0, \pi]$ and differentiable on $(0, \pi)$.

Also,

$f(\pi)=f(0)=0$

Thus, $f(x)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c \in(0, \pi)$ such that $f^{\prime}(c)=0$.

We have

$f(x)=\frac{\sin x}{e^{x}}$

$\Rightarrow f^{\prime}(x)=\frac{\cos x-\sin x}{e^{x}}$

$\therefore f^{\prime}(x)=0$

$\Rightarrow \frac{\cos x-\sin x}{e^{x}}=0$

$\Rightarrow \cos x-\sin x=0$

$\Rightarrow \tan x=1$

 

$\Rightarrow x=\frac{\pi}{4}$

Thus, $c=\frac{\pi}{4} \in(0, \pi)$ such that $f^{\prime}(c)=0$.

​Hence, Rolle's theorem is verified.

(viii)

The given function is $f(x)=\sin 3 x$.

Since $\sin 3 x$ is everywhere continuous and differentiable, $\sin 3 x$ is continuous on $[0, \pi]$ and differentiable on $(0, \pi)$.

Also,

$f(\pi)=f(0)=0$

Thus, $f(x)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c \in(0, \pi)$ such that $f^{\prime}(c)=0$.

We have

$f(x)=\sin 3 x$

 

$\Rightarrow f^{\prime}(x)=3 \cos 3 x$

$\therefore f^{\prime}(x)=0$

$\Rightarrow 3 \cos 3 x=0$

$\Rightarrow \cos 3 x=0$

$\Rightarrow 3 x=\frac{\pi}{2}, \frac{3 \pi}{2}, \ldots$

 

$\Rightarrow x=\frac{\pi}{6}, \frac{\pi}{2}, \frac{5 \pi}{6}$

Thus, $c=\frac{\pi}{6}, \frac{\pi}{2}, \frac{5 \pi}{6} \in(0, \pi)$ such that $f^{\prime}(c)=0$

​Hence, Rolle's theorem is verified.

(ix)

The given function is $f(x)=e^{1-x^{2}}$.

Since exponential function is everywhere continuous and differentiable, $e^{1-x^{2}}$ is continuous on $[-1,1]$ and differentiable on $(-1,1)$

Also,

$f(1)=f(-1)=1$

Thus, $f(x)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c \in(-1,1)$ such that $f^{\prime}(c)=0$.

We have

$f(x)=e^{1-x^{2}}$

$\Rightarrow f^{\prime}(x)=-2 x e^{1-x^{2}}$

$\therefore f^{\prime}(x)=0$

$\Rightarrow-2 x e^{1-x^{2}}=0$

 

$\Rightarrow x=0$

Thus, $c=0 \in(-1,1)$ such that $f^{\prime}(c)=0$.

Hence, Rolle's theorem is verified.

(x)

The given function is $f(x)=\log \left(x^{2}+2\right)-\log 3$, which can be rewritten as $f(x)=\log \left(\frac{x^{2}+2}{3}\right)$.

Since logarithmic function is differentiable and so continuous in its domain, $f(x)=\log \left(\frac{x^{2}+2}{3}\right)$ is continuous on $[-1,1]$ and differentiable on $(-1,1)$.

Also,

$f(1)=f(-1)=0$

Thus, $f(x)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c \in(-1,1)$ such that $f^{\prime}(c)=0$.

We have

$f(x)=\log \left(\frac{x^{2}+2}{3}\right)$

$\Rightarrow f^{\prime}(x)=\frac{3(2 x)}{x^{2}+2}=\frac{6 x}{x^{2}+2}$

$\therefore f^{\prime}(x)=0$

$\Rightarrow \frac{6 x}{x^{2}+2}=0$

$\Rightarrow x=0$

Thus, $c=0 \in(-1,1)$ such that $f^{\prime}(c)=0$.

​Hence, Rolle's theorem is verified.

(xi)

The given function is $f(x)=\sin x+\cos x$.

Since $\sin x$ and $\cos x$ are everywhere continuous and differentiable, $f(x)=\sin x+\cos x$ is continuous on $\left[0, \frac{\pi}{2}\right]$ and differentiable on $\left(0, \frac{\pi}{2}\right)$.

Also,

$f\left(\frac{\pi}{2}\right)=f(0)=1$

Thus, $f(x)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c \in\left(0, \frac{\pi}{2}\right)$ such that $f^{\prime}(c)=0$.

We have

$f(x)=\sin x+\cos x$

 

$\Rightarrow f^{\prime}(x)=\cos x-\sin x$

$\therefore f^{\prime}(x)=0$

$\Rightarrow \cos x-\sin x=0$

$\Rightarrow \tan x=1$

 

$\Rightarrow x=\frac{\pi}{4}$

Thus, $c=\frac{\pi}{4} \in\left(0, \frac{\pi}{2}\right)$ such that $f^{\prime}(c)=0$.

Hence, Rolle's theorem is verified.

(xii)

The given function is $f(x)=2 \sin x+\sin 2 x$.

Since $\sin x \& \sin 2 x$ are everywhere continuous and differentiable, $f(x)$ is continuous on $[0, \pi]$ and differentiable on $(0, \pi)$.

Also,

$f(\pi)=f(0)=0$

Thus, $f(x)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c \in(0, \pi)$ such that $f^{\prime}(c)=0$.

We have

$f(x)=2 \sin x+\sin 2 x$

$\Rightarrow f^{\prime}(x)=2 \cos x+2 \cos 2 x$

$\therefore f^{\prime}(x)=0$

$\Rightarrow 2 \cos x+2 \cos 2 x=0$

$\Rightarrow \cos x+\cos 2 x=0$

$\Rightarrow \cos x+2 \cos ^{2} x-1=0$

$\Rightarrow 2 \cos ^{2} x+\cos x-1=0$

$\Rightarrow(\cos x+1)(2 \cos x-1)=0$

$\Rightarrow \cos x=-1, \cos x=\frac{1}{2}$

$\Rightarrow \cos x=\cos \pi, \cos x=\frac{\pi}{3}$

 

$\Rightarrow x=\pi, \frac{\pi}{3}$

Thus, $c=\frac{\pi}{3} \in(0, \pi)$ such that $f^{\prime}(c)=0$.

​Hence, Rolle's theorem is verified

(xiii)

The given function is $f(x)=\frac{x}{2}-\sin \frac{\pi x}{6}$.

Since $\sin x \& \frac{x}{2}$ are everywhere continuous and differentiable, $f(x)$ is continuous on $[-1,0]$ and differentiable on $(-1,0)$.

Also,

$f(-1)=f(0)=0$

Thus, $f(x)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c \in(-1,0)$ such that $f^{\prime}(c)=0$.

We have

$f(x)=\frac{x}{2}-\sin \frac{\pi x}{6}$

 

$\Rightarrow f^{\prime}(x)=\frac{1}{2}-\frac{\pi}{6} \cos \frac{\pi x}{6}$

$\therefore f^{\prime}(x)=0$

$\Rightarrow \frac{1}{2}-\frac{\pi}{6} \cos \frac{\pi x}{6}=0$

$\Rightarrow \cos \frac{\pi x}{6}=\frac{3}{\pi}$

 

$\Rightarrow x=\frac{-6}{\pi} \cos ^{-1}\left(\frac{3}{\pi}\right)$

Thus, $c=\frac{-6}{\pi} \cos ^{-1}\left(\frac{3}{\pi}\right) \in(-1,0)$ such that $f^{\prime}(c)=0$.

Hence, Rolle's theorem is verified.

(xiv)

The given function is $f(x)=\frac{6 x}{\pi}-4 \sin ^{2} x$.

Since $\sin ^{2} x \& x$ are everywhere continuous and differentiable, $f(x)$ is continuous on $\left[0, \frac{\pi}{6}\right]$ and differentiable on $\left(0, \frac{\pi}{6}\right)$.

Also,

$f\left(\frac{\pi}{6}\right)=f(0)=0$

Thus, $f(x)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c \in\left(0, \frac{\pi}{6}\right)$ such that $f^{\prime}(c)=0$.

We have

$f(x)=\frac{6 x}{\pi}-4 \sin ^{2} x$

 

$\Rightarrow f^{\prime}(x)=\frac{6}{\pi}-8 \sin x \cos x$

$\therefore f^{\prime}(x)=0$

$\Rightarrow \frac{6}{\pi}-8 \sin x \cos x=0$

$\Rightarrow \sin 2 \mathrm{x}=\frac{3}{2 \pi}$

 

$\Rightarrow x=\frac{1}{2} \sin ^{-1}\left(\frac{3}{2 \pi}\right)$

Thus, $c=\frac{1}{2} \sin ^{-1}\left(\frac{3}{2 \pi}\right) \in\left(0, \frac{\pi}{6}\right)$ such that $f^{\prime}(c)=0$.

​Hence, Rolle's theorem is verified.

(xv)

The given function is $f(x)=4^{\sin x}$.

Since sine function and exponential function are everywhere continuous and differentiable, $f(x)$ is continuous on $[0, \pi]$ and differentiable on $(0, \pi)$.

Also,

$f(\pi)=f(0)=1$

Thus, $f(x)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c \in(0, \pi)$ such that $f^{\prime}(c)=0$.

We have

$f(x)=4^{\sin x}$

 

$\Rightarrow f^{\prime}(x)=4^{\sin x}(\cos x) \log 4$

$\therefore f^{\prime}(x)=0$

$\Rightarrow 4^{\sin x}(\cos x) \log 4=0$

$\Rightarrow 4^{\sin x} \cos x=0$

$\Rightarrow \cos x=0$

 

$\Rightarrow x=\frac{\pi}{2}$

Thus, $c=\frac{\pi}{2} \in(0, \pi)$ such that $f^{\prime}(c)=0$.

Hence, Rolle's theorem is verified.

(xvi)  

According to Rolle's theorem, if $f(x)$ is a real valued function defined on $[a, b]$ such that it is continuous on $[a, b]$, it is differentiable on $(a, b)$ and $f(a)=f(b)$, then there exists a real number $c \in(a, b)$ such that $f(c)=0$.

Now, $f(x)$ is defined for all $x \in[1,4]$.

 

At each point of $[1,4]$, the limit of $f(x)$ is equal to the value of the function. Therefore, $f(x)$ is continuous on $[1,4]$.

Also, $f^{\prime}(x)=2 x-5$ exists for all $x \in(1,4)$.

So, $f(x)$ is differentiable on $(1,4)$.

Also,

$f(1)=f(4)=0$

Thus, all the three conditions of Rolle’s theorem are satisfied.

Now, we have to show that there exists $c \in(1,4)$ such that $f^{\prime}(c)=0$.

We have

$f^{\prime}(x)=2 x-5$

Thus, all the three conditions of Rolle’s theorem are satisfied.

Now, we have to show that there exists $c \in(1,4)$ such that $f^{\prime}(c)=0$.

We have

$f^{\prime}(x)=2 x-5$

$\therefore f^{\prime}(x)=0$

$\Rightarrow 2 x-5=0$

 

$\Rightarrow x=\frac{5}{2}$

[Since $c=\frac{5}{2} \in(1,4)$ such that $f^{\prime}(c)=0$ ]

Hence, Rolle’s theorem is verified.

(xvii)

The given function is $f(x)=\sin ^{4} x+\cos ^{4} x$.

Since $\sin x$ and $\cos x$ are everywhere continuous and differentiable, $f(x)=\sin ^{4} x+\cos ^{4} x$ is continuous on $\left[0, \frac{\pi}{2}\right]$ and differentiable on $\left(0, \frac{\pi}{2}\right)$.

Also,

$f\left(\frac{\pi}{2}\right)=f(0)=1$

Thus, $f(x)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c \in\left(0, \frac{\pi}{2}\right)$ such that $f^{\prime}(c)=0$.

We have

$f(x)=\sin ^{4} x+\cos ^{4} x$

 

$\Rightarrow f^{\prime}(x)=4 \sin ^{3} x \cos x-4 \cos ^{3} x \sin x$

$\therefore f^{\prime}(x)=0$

$\Rightarrow 4 \sin ^{3} x \cos x-4 \cos ^{3} x \sin x=0$

$\Rightarrow \sin ^{3} x \cos x-\cos ^{3} x \sin x=0$

$\Rightarrow \tan ^{3} x-\tan x=0$

$\Rightarrow \tan x\left(\tan ^{2} x-1\right)=0$

$\Rightarrow \tan x=0, \tan ^{2} x=1$

$\Rightarrow \tan x=0, \tan x=\pm 1$

 

$\Rightarrow x=0, x=\frac{\pi}{4}, \frac{3 \pi}{4}$

Thus, $c=\frac{\pi}{4} \in\left(0, \frac{\pi}{2}\right)$ such that $f^{\prime}(c)=0$.

​Hence, Rolle's theorem is verified.

(xviii)

The given function is $f(x)=\sin x-\sin 2 x$.

Since $\sin x$ and $\sin 2 x$ are everywhere continuous and differentiable, $f(x)$ is continuous on $[0, \pi]$ and differentiable on $(0, \pi)$.

Also,

$f(\pi)=f(0)=0$

Thus, $f(x)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists $c \in(0, \pi)$ such that $f^{\prime}(c)=0$.

We have

$f(x)=\sin x-\sin 2 x$

 

$\Rightarrow f^{\prime}(x)=\cos x-2 \cos 2 x$

$\therefore f^{\prime}(x)=0$

$\Rightarrow \cos x-2 \cos 2 x=0$

$\Rightarrow \cos x-2 \cos ^{2} x+1=0$

$\Rightarrow 2 \cos ^{2} x-\cos x-1=0$

$\Rightarrow(\cos x-1)(2 \cos x+1)=0$

$\Rightarrow \cos x=1, \cos x=\frac{-1}{2}$

$\Rightarrow \cos x=\cos \frac{\pi}{2}, \cos x=\frac{2 \pi}{3}$

 

$\Rightarrow x=\frac{\pi}{2}, \frac{2 \pi}{3}$

Thus, $c=\frac{\pi}{2}, \frac{2 \pi}{3} \in(0, \pi)$ such that $f^{\prime}(c)=0$.

Hence, Rolle's theorem is verified.​