Question
Verify Rolle's theorem for the following function on the indicated intervals

f(x) = x(x - 2)2 on the interval [0, 2]

Answer

Given function is f(x) = x(x - 2)2. Which can be rewritten as f(x) = x3 - 4x2 + 4x.
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, 2].
Also,
f(0) = f(2) = 0
Thus, all the conditions of Rolle's theorem are satisfied.
Now, we have to show that there exists $\text{c}\in[0,2]$ such that f'(c) = 0.
We have
f(x) = x3 - 4x2 + 4x
⇒ f'(x) = 3x- 8x + 4
When, f'(x) = 0
3x- 8x + 4 = 0
⇒ 3x- 6x - 2x + 4 = 0
⇒ 3x(x - 2) - 2(x - 2) = 0
⇒ (x - 2)(3x - 2)
$\Rightarrow\text{x}=2,\frac{2}{3}$
Thus, $\text{c}=\frac{2}{3}\in(0,2)$ such that f'(c) = 0.
Hence, Rolle's theorem is verified.

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