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,
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