Question

Verify Rolle's theorem for the function f(x) = x(x − 4)² on the interval [0, 4].

Answer 1

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

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 $\left[0,4\right]$.

Also,
$f\left(0\right)=f\left(4\right)=0$

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

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

We have
$$\begin{gathered} f\left(x\right)=x^3-8x^2+16x \\ \Rightarrow f\text{'}\left(x\right)=3x^2-16x+16 \\ \therefore f\text{'}\left(x\right)=0 \\ \Rightarrow 3x^2-16x+16=0 \\ \Rightarrow 3x^2-12x-4x+16=0 \\ \Rightarrow 3x\left(x-4\right)-4\left(x-4\right)=0 \\ \Rightarrow \left(x-4\right)\left(3x-4\right) \\ \Rightarrow x=4,\frac{4}{3} \end{gathered}$$

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

Hence, Rolle's theorem is verified.