Question

If $a,b$ are two distinct real numbers $(a<b)$ in the domain of $f\left(x\right)=4x^3+3x^2-x-1,$ then the minimum value of $\frac{f\left(b\right)-f\left(a\right)}{b-a}$ is

• A. $-\frac{7}{4}$
• B. $-\frac{7}{2}$
• C. $\frac{7}{2}$
• D. $\frac{7}{4}$

Answer 1

The correct option is A $-\frac{7}{4}$

$f\left(x\right)=4x^3+3x^2-x-1,$ is continuous and differentiable in $[a,b]$
So, there exists $x=c$ such that
$f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$ (from LMVT)
$\Rightarrow \frac{f\left(b\right)-f\left(a\right)}{b-a}=12c^2+6c-1$
$=12{(c+\frac{1}{4})}^2-\frac{7}{4}\ge -\frac{7}{4}$