Question
Verify Rolle's theorem for the following function on the indicated intervals
f(x) = (x2 - 1)(x - 2) on [-1, 2]
f(x) = (x2 - 1)(x - 2) on [-1, 2]
f(x) = (x2 - 1)(x - 2) on [-1, 2]
f(x) is continuous is [-1, 2] and differentiable in (-1, 2) as it is a polynomial functions.
Now,
f(-1) = (1-1)(-1-2) = 0
f(2) = (4-1)(2-2) = 0
⇒ f(-1) = f(2)
So, Rolle's theorem is applicable on f(x) is [-1, 2] therefore, we have to show that there exist a $\text{c}\in(-1,2)$ such that f'(c) = 0
Now,
f(x) = (x2 - 1)(x - 2)
f'(x) = 2x(x - 2) + (x2 - 1)
= 2x2 - 4 + x2 - 1
f'(x) = 3x2 - 5
Now,
f'(c) = 0
⇒ 3x2 - 5 = 0
$\Rightarrow\text{x}=-\sqrt{\frac{5}{3}}$ or $\text{x}=\sqrt{\frac{5}{3}}\in(-1,2)$
Thus, Rolle's theorem is verified.
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$\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}},\ 4\vec{\text{a}}+3\vec{\text{b}},\ 10\vec{\text{a}}+7\vec{\text{b}}-2\vec{\text{c}}$