Question
Verify Rolle's theorem for the following function on the indicated intervals
f(x) = x2 - 8x + 12 on [2, 6]
f(x) = x2 - 8x + 12 on [2, 6]
We know that a polynomial function is everywhere derivable and hence continuous.
So, being a polynomial function f(x) is continuous and derivable on [2, 6].
f(2) = (2)2 - 8(2) + 12 = 4 - 16 + 12 = 0
f(6) = (6)2 - 8(6) + 12 = 36 - 48 + 12 = 0
$\therefore$ f(2) = f(6) = 0
Thus, all the conditions of rolle's theorem are satisfied.
Now, we have to show that there exist
$\text{c}\in(2, 6)$ such that f'(c) = 0We have
f(x) = x2 - 8x + 12
⇒ f'(x) = 2x - 8
$\therefore$
f'(x) = 0⇒ 2x - 8 = 0
⇒ x = 4
Thus,
$\text{c}=4\in(2,6)$ such that f'(c) = 0Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\text{x}^{2}\frac{\text{dy}}{\text{dx}}=\text{y}^{2}+\text{2xy}$
Given that y = 1, when x = 1.