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

3 $\frac{5}{2}$ $\frac{9}{4}$ $\frac{7}{2}$

3

$f(x)=x^2-6 x+8,2 \leq x \leq 4$ f(2) = 0 = f(4) ∴ By Rolle's theorme, ∃ c ∈ (2, 4) such that f'(c) = 0 ⇒  2c – 6 = 0 ⇒ c = 3