For the function $f(x)=x^2-6 x+8,2 \leq x \leq 4$, the value of x for which f'(x) vanishes is
Answer & explanation
Correct answer: option 1
$f(x)=x^2-6 x+8,2 \leq x \leq 4$
f(2) = 0 = f(4)
∴ By Rolle's theorme, ∃ c ∈ (2, 4)
such that f'(c) = 0
⇒ 2c – 6 = 0 ⇒ c = 3