Question

Verify Rolle's theorem for each of the following functions on the indicated intervals
(i) f(x) = x² − 8x + 12 on [2, 6]

(ii) f(x) = x² − 4x + 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) = (x² − 1) (x − 2) on [−1, 2]

Answer 1

(i) Given:
$f\left(x\right)=x^2-8x+12$

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

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

Also,
$$\begin{gathered} f\left(2\right)={\left(2\right)}^2-8\left(2\right)+12=4-16+12=0 \\ f\left(6\right)={\left(6\right)}^2-8\left(6\right)+12=36-48+12=0 \\ \therefore f\left(2\right)=f\left(6\right)=0 \end{gathered}$$

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

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

We have
$$\begin{gathered} f\left(x\right)=x^2-8x+12 \\ \Rightarrow f\text{'}\left(x\right)=2x-8 \\ \therefore f\text{'}\left(x\right)=0\Rightarrow 2x-8=0\Rightarrow x=4 \end{gathered}$$

Thus, $c=4\in \left(2,6\right)\mathrm{such}\mathrm{that}f\text{'}\left(c\right)=0$.

Hence, Rolle's theorem is verified.

(ii) Given:
$f\left(x\right)=x^2-4x+3$

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

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

Also,
$$\begin{gathered} f\left(1\right)={\left(1\right)}^2-4\left(1\right)+3=1-4+3=0 \\ f\left(3\right)={\left(3\right)}^2-4\left(3\right)+3=9-12+3=0 \\ \therefore f\left(1\right)=f\left(3\right)=0 \end{gathered}$$

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

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

We have
$$\begin{gathered} f\left(x\right)=x^2-4x+3 \\ \Rightarrow f\text{'}\left(x\right)=2x-4 \\ \therefore f\text{'}\left(x\right)=0\Rightarrow 2x-4=0\Rightarrow x=2 \end{gathered}$$

Thus, $c=2\in \left(1,3\right)\mathrm{such}\mathrm{that}f\text{'}\left(c\right)=0$.

Hence, Rolle's theorem is verified.

(iii) Given:
$f\left(x\right)=\left(x-1\right){\left(x-2\right)}^2$
i.e. $f\left(x\right)=x^3+4x-4x^2-x^2-4+4x$
i.e. $f\left(x\right)=x^3-5x^2+8x-4$

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

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

Also,
$$\begin{gathered} f\left(1\right)={\left(1\right)}^3-5{\left(1\right)}^2+8\left(1\right)-4=0 \\ f\left(2\right)={\left(2\right)}^3-5{\left(2\right)}^2+8\left(2\right)-4=0 \\ \therefore f\left(1\right)=f\left(2\right)=0 \end{gathered}$$

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

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

We have
$$\begin{gathered} f\left(x\right)=x^3+8x-5x^2-4 \\ \Rightarrow f\text{'}\left(x\right)=3x^2+8-10x \\ \therefore f\text{'}\left(x\right)=0\Rightarrow 3x^2-10x+8=0 \\ \Rightarrow 3x^2-6x-4x+8=0 \\ \Rightarrow 3x\left(x-2\right)-4\left(x-2\right)=0 \\ \Rightarrow \left(x-2\right)\left(3x-4\right) \\ \Rightarrow x=2,\frac{4}{3} \end{gathered}$$

Thus, $c=\frac{4}{3}\in \left(1,2\right)\mathrm{such}\mathrm{that}f\text{'}\left(c\right)=0$.

Hence, Rolle's theorem is verified.

(iv) Given:
$f\left(x\right)=x{\left(x-1\right)}^2$
$\Rightarrow f\left(x\right)=x\left(x^2-2x+1\right)$
$\therefore f\left(x\right)=\left(x^3-2x^2+x\right)$

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

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

Also,
$f\left(0\right)=f\left(1\right)=0$

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

Now, we have to show that there exists $c\in \left(0,1\right)$ such that $f\text{'}\left(c\right)=0$.

We have
$$\begin{gathered} f\left(x\right)=x^3-2x^2+x \\ \Rightarrow f\text{'}\left(x\right)=3x^2-4x+1 \\ \therefore f\text{'}\left(x\right)=0\Rightarrow 3x^2-4x+1=0 \\ \Rightarrow 3x^2-3x-x+1=0 \\ \Rightarrow 3x\left(x-1\right)-1\left(x-1\right)=0 \\ \Rightarrow \left(x-1\right)\left(3x-1\right)=0 \\ \Rightarrow x=1,\frac{1}{3} \end{gathered}$$

Thus, $c=\frac{1}{3}\in \left(0,1\right)\mathrm{such}\mathrm{that}f\text{'}\left(c\right)=0$.

Hence, Rolle's theorem is verified.

(v) Given:
$f\left(x\right)=\left(x^2-1\right)\left(x-2\right)$
i.e. $f\left(x\right)=x^3-2x^2-x+2$

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

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

Also,
$f\left(-1\right)=f\left(2\right)=0$

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

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

We have
$$\begin{gathered} f\left(x\right)=x^3-x-2x^2+2 \\ \Rightarrow f\text{'}\left(x\right)=3x^2-4x-1 \\ \therefore f\text{'}\left(x\right)=0\Rightarrow 3x^2-4x-1=0 \\ \Rightarrow x=\frac{-\left(-4\right)\pm \sqrt{{\left(-4\right)}^2-4\times 3\times \left(-1\right)}}{2\times 3} \\ \Rightarrow x=\frac{4\pm \sqrt{16+12}}{6} \\ \Rightarrow x=\frac{4\pm \sqrt{28}}{6} \\ \Rightarrow x=\frac{4\pm 2\sqrt{7}}{6} \\ \Rightarrow x=\frac{2\pm \sqrt{7}}{3} \\ \Rightarrow x=\frac{1}{3}\left(2-\sqrt{7}\right),\frac{1}{3}\left(2+\sqrt{7}\right) \end{gathered}$$

Thus, $c=\frac{1}{3}\left(2-\sqrt{7}\right),\frac{1}{3}\left(2+\sqrt{7}\right)\in \left(-1,2\right)\mathrm{such}\mathrm{that}f\text{'}\left(c\right)=0$.

Hence, Rolle's theorem is verified.