Question

If Rolle's theorem holds for the function $f\left(x\right)=2x^3+b{x}^2+cx,x\in [–1,1],$ at the point $x=\frac{1}{2}$, then $8b+c$ is

• A. 2.000
• B. 2
• C. 2.0
• D. 2.00

Answer 1

Answer: (A), (B), (C), (D)

By Rolle's theorem
$f\left(1\right)=f(-1)2+b+c=–2+b–c\Rightarrow c=–2f^{\prime }\left(x\right)=6x^2+2bx+c{f}^{\prime }\left(\frac{1}{2}\right)=\frac{3}{2}+b+c=0\Rightarrow b=\frac{1}{2}\text{So,}(8b+c)=2$