**Question:**

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

(i) $f(x)=x^{2}-8 x+12$ on $[2,6]$

(ii) $f(x)=x^{2}-4 x+3$ on $[1,3]$

(iii) $f(x)=(x-1)(x-2)^{2}$ on $[1,2]$

(iv) $f(x)=x(x-1)^{2}$ on $[0,1]$

(v) $f(x)=\left(x^{2}-1\right)(x-2)$ on $[-1,2]$

(vi) $f(x)=x(x-4)^{2}$ on the interval $[0,4]$

(vii) $f(x)=x(x-2)^{2}$ on the interval $[0,2]$

(viii) $f(x)=x^{2}+5 x+6$ on the interval $[-3,-2]$

**Solution:**

(i) Given:

$f(x)=x^{2}-8 x+12$

We know that a polynomial function is everywhere derivable and hence continuous.

So, being a polynomial function, $f(x)$ is continuous and derivable on $[2,6]$.

Also,

$f(2)=(2)^{2}-8(2)+12=4-16+12=0$

$f(6)=(6)^{2}-8(6)+12=36-48+12=0$

$\therefore f(2)=f(6)=0$

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

Now, we have to show that there exists $c \in(2,6)$ such that $f^{\prime}(c)=0$.

We have

$f(x)=x^{2}-8 x+12$

$\Rightarrow f^{\prime}(x)=2 x-8$

$\therefore f^{\prime}(x)=0 \Rightarrow 2 x-8=0 \Rightarrow x=4$

Thus, $c=4 \in(2,6)$ such that $f^{\prime}(c)=0$

Hence, Rolle's theorem is verified.

(ii) Given:

$f(x)=x^{2}-4 x+3$

We know that a polynomial function is everywhere derivable and hence continuous.

So, being a polynomial function, $f(x)$ is continuous and derivable on $[1,3]$.

Also,

$f(1)=(1)^{2}-4(1)+3=1-4+3=0$

$f(3)=(3)^{2}-4(3)+3=9-12+3=0$

$\therefore f(1)=f(3)=0$

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

Now, we have to show that there exists $c \in(1,3)$ such that $f^{\prime}(c)=0$.

We have

$f(x)=x^{2}-4 x+3$

$\Rightarrow f^{\prime}(x)=2 x-4$

$\therefore f^{\prime}(x)=0 \Rightarrow 2 x-4=0 \Rightarrow x=2$

Thus, $c=2 \in(1,3)$ such that $f^{\prime}(c)=0$.

Hence, Rolle's theorem is verified.

(iii) Given:

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

i.e. $f(x)=x^{3}+4 x-4 x^{2}-x^{2}-4+4 x$

i.e. $f(x)=x^{3}-5 x^{2}+8 x-4$

We know that a polynomial function is everywhere derivable and hence continuous.

So, being a polynomial function, $f(x)$ is continuous and derivable on $[1,2]$.

Also,

$f(1)=(1)^{3}-5(1)^{2}+8(1)-4=0$

$f(2)=(2)^{3}-5(2)^{2}+8(2)-4=0$

$\therefore f(1)=f(2)=0$

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

Now, we have to show that there exists $c \in(1,2)$ such that $f^{\prime}(c)=0$.

We have

$f(x)=x^{3}+8 x-5 x^{2}-4$

$\Rightarrow f^{\prime}(x)=3 x^{2}+8-10 x$

$\therefore f^{\prime}(x)=0 \Rightarrow 3 x^{2}-10 x+8=0$

$\Rightarrow 3 x^{2}-6 x-4 x+8=0$

$\Rightarrow 3 x(x-2)-4(x-2)=0$

$\Rightarrow(x-2)(3 x-4)$

$\Rightarrow x=2, \frac{4}{3}$

Thus, $c=\frac{4}{3} \in(1,2)$ such that $f^{\prime}(c)=0$.

Hence, Rolle's theorem is verified.

(iv) Given:

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

$\Rightarrow f(x)=x\left(x^{2}-2 x+1\right)$

$\therefore f(x)=\left(x^{3}-2 x^{2}+x\right)$

We know that a polynomial function is everywhere derivable and hence continuous.

So, $f(x)$ being a polynomial function is continuous and derivable on $[0,1]$.

Also,

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

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

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

We have

$f(x)=x^{3}-2 x^{2}+x$

$\Rightarrow f^{\prime}(x)=3 x^{2}-4 x+1$

$\therefore f^{\prime}(x)=0 \Rightarrow 3 x^{2}-4 x+1=0$

$\Rightarrow 3 x^{2}-3 x-x+1=0$

$\Rightarrow 3 x(x-1)-1(x-1)=0$

$\Rightarrow(x-1)(3 x-1)=0$

$\Rightarrow x=1, \frac{1}{3}$

Thus, $c=\frac{1}{3} \in(0,1)$ such that $f^{\prime}(c)=0$.

Hence, Rolle's theorem is verified.

(v) Given:

$f(x)=\left(x^{2}-1\right)(x-2)$

i.e. $f(x)=x^{3}-2 x^{2}-x+2$

We know that a polynomial function is everywhere derivable and hence continuous.

So, being a polynomial function, $f(x)$ is continuous and derivable on $[-1,2]$.'

Also,

$f(-1)=f(2)=0$

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

Now, we have to show that there exists $c \in(-1,2)$ such that $f^{\prime}(c)=0$.

We have

$f(x)=x^{3}-x-2 x^{2}+2$

$\Rightarrow f^{\prime}(x)=3 x^{2}-4 x-1$

$\therefore f^{\prime}(x)=0 \Rightarrow 3 x^{2}-4 x-1=0$

$\Rightarrow x=\frac{-(-4) \pm \sqrt{(-4)^{2}-4 \times 3 \times(-1)}}{2 \times 3}$

$\Rightarrow x=\frac{4 \pm \sqrt{16+12}}{6}$

$\Rightarrow x=\frac{4 \pm \sqrt{28}}{6}$

$\Rightarrow x=\frac{4 \pm 2 \sqrt{7}}{6}$

$\Rightarrow x=\frac{2 \pm \sqrt{7}}{3}$

$\Rightarrow x=\frac{1}{3}(2-\sqrt{7}), \frac{1}{3}(2+\sqrt{7})$

Thus, $c=\frac{1}{3}(2-\sqrt{7}), \frac{1}{3}(2+\sqrt{7}) \in(-1,2)$ such that $f^{\prime}(c)=0$

Hence, Rolle's theorem is verified.

(vi) Given function is $f(x)=x(x-4)^{2}$, which can be rewritten as $f(x)=x^{3}-8 x^{2}+16 x$.

We know that a polynomial function is everywhere derivable and hence continuous.

So, being a polynomial function, $f(x)$ is continuous and derivable on $[0,4]$.

Also,

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

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

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

We have

$f(x)=x^{3}-8 x^{2}+16 x$

$\Rightarrow f^{\prime}(x)=3 x^{2}-16 x+16$

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

$\Rightarrow 3 x^{2}-16 x+16=0$

$\Rightarrow 3 x^{2}-12 x-4 x+16=0$

$\Rightarrow 3 x(x-4)-4(x-4)=0$

$\Rightarrow(x-4)(3 x-4)$

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

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

Hence, Rolle's theorem is verified.

(vii) The given function is $f(x)=x(x-2)^{2}$, which can be rewritten as $f(x)=x^{3}-4 x^{2}+4 x$.

We know that a polynomial function is everywhere derivable and hence continuous.

So, being a polynomial function, $f(x)$ is continuous and derivable on $[0,2]$.

Also,

$f(0)=f(2)=0$

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

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

We have

$f(x)=x^{3}-4 x^{2}+4 x$

$\Rightarrow f^{\prime}(x)=3 x^{2}-8 x+4$

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

$3 x^{2}-8 x+4=0$

$\Rightarrow 3 x^{2}-6 x-2 x+4=0$

$\Rightarrow 3 x(x-2)-2(x-2)=0$

$\Rightarrow(x-2)(3 x-2)$

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

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

Hence, Rolle's theorem is verified.

(viii) Given function is $f(x)=x^{2}+5 x+6$.

We know that a polynomial function is everywhere derivable and hence continuous.

So, being a polynomial function, $f(x)$ is continuous and derivable on $[-3,-2]$.

Also,

$f(-3)=(-3)^{2}+5(-3)+6=9-15+6=0$

$f(-2)=(-2)^{2}+5(-2)+6=4-10+6=0$

$\therefore f(-3)=f(-2)=0$

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

Now, we have to show that there exists $c \in[-3,-2]$ such that $f^{\prime}(c)=0$.

We have

$f(x)=x^{2}+5 x+6$

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

$\therefore f^{\prime}(x)=0 \Rightarrow 2 x+5=0$

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

Thus, $c=\frac{-5}{2} \in(-3,-2)$ such that $f^{\prime}(c)=0$

Hence, Rolle's theorem is verified.