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) = x3 - 8x2 + 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) = x3 - 8x2 + 16x
⇒ f'(x) = 3x- 16x + 16
$\therefore$ f'(x) = 0
⇒ 3x- 16x + 16 = 0
⇒ 3x- 12x - 4x + 16 = 0
⇒ 3x(x - 4) - 4(x - 4) = 0
⇒ (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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