If Rolle's theorem holds for the function f(x) = 2x3 + bx2 + cx, x ϵ [-1, 1], at the point x = (1/2), then 2b + c equals :

Question

If Rolle's theorem holds for the function f(x) = 2x3 + bx2 + cx, x ϵ [-1, 1], at the point x = (1/2), then 2b + c equals : (A) 1 (B) -1 (C) 2 (D) -3

Tardigrade Admin · Accepted Answer

f(-1)=f(1) and f'((1/2))=0 ⇒ -2+b-c=2+b+c ⇒ c=-2 and f'(x)=6 x2+2 b x +c ⇒ f'((1/2))=(3/2)+b-2=0 ⇒ b=(1/2) ⇒ 2 b +c=-1

Q. If Rolle's theorem holds for the function $f (x) = 2 x^{3} + b x^{2} + c x, x ϵ [- 1, 1],$ at the point $x = \frac{1}{2},$ then $2 b + c$ equals :