Question
Verify Rolle's theorem for the following function on the indicated intervals$f(x) = x(x^- 4)^2$ on the interval $[0, 4]$

Answer

Given function is $f(x) = x(x^- 4)^2$. Which can be rewritten as $f(x) = x^3 - 8x^2 + 16x.$
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 $\text{c}\in[0,4]$ such that $f'(c) = 0.$
We have
$f(x) = x^3- 8x^2 + 16x$
$\Rightarrow f'(x) = 3x^2- 16x + 16$
$\therefore f'(x) = 0$
$\Rightarrow 3x^2- 16x + 16 = 0$
$\Rightarrow 3x^2- 12x - 4x + 16 = 0$
$\Rightarrow 3x(x - 4) - 4(x - 4) = 0$
$\Rightarrow (x - 4)(3x - 4)$
$\Rightarrow\text{x}=4,\frac{4}{3}$
Thus, $\text{c}=\frac{4}{3}\in(0,4)$ such that $f'(c) = 0.$
Hence, Rolle's theorem is verified.

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