# Verify Rolle's theorem for the following function; $f(x)=4x^{3}-9x;-\large\frac{3}{2}\leq x\leq \frac{3}{2}$.

Note: This is part 4th of a 4 part question, split as 4 separate questions here.

Toolbox:
• Rolle's Theorem: Let $f$ be a real valued function that satisfies the following conditions:
• (i) f is defined and continuous on the closed interval $[a,b]$
• (ii) f is differentiable in the open interval $(a,b)$
• (iii) $f(a)=f(b)$
• Then there exists at least one value $c \in (a,b)$ such that $f'(c)=0$
$f(x)=4x^3-9x, \qquad \large\frac{-3}{2}$$\leq x \leq \large\frac{3}{2} Step 1: f(x) os a polynomial function. So it is continuous in \bigg[\large\frac{-3}{2},\frac{3}{2}\bigg] and differentiable in \bigg(\large\frac{-3}{2},\frac{3}{2}\bigg) f\bigg(\large\frac{-3}{2}\bigg)$$ =4 \bigg(\large\frac{-3}{2}\bigg)^3$$-9 \bigg(\large\frac{-3}{2} \bigg)=\frac{-27}{2}+\frac{27}{2}$$=0$
$f\bigg(\large\frac{3}{2}\bigg)$$=4 \bigg(\large\frac{3}{2}\bigg)^3$$-9 \bigg(\large\frac{3}{2} \bigg)=\frac{27}{2}-\frac{27}{2}$$=0 f\bigg(\large\frac{-3}{2}\bigg)=f\bigg(\large\frac{3}{2}\bigg)$$=0$
The conditions for Roller's theorem are satisfied
Step 2:
$f'(x)=12 x^2-9$
$f'(c)=12c^2-9 \qquad f'(c)=0$
=>$12 c^2-9=0 => c= \pm \large\frac{\sqrt 3}{2}$
Both these values are in $\bigg(\large\frac{-3}{2},\frac{3}{2}\bigg)$
Hence Rolle's theorem is verified