Mean Value Theorems — Maths STD 12 Science — Question

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.