Question

The value of $c$ in the Lagrange's mean value theorem for the function $f\left(x\right)=x^3-4x^2+8x+11$, where $x\in [0,1]$ is :

• A. $\frac{4-\sqrt{7}}{3}$
• B. $\frac{2}{3}$
• C. $\frac{\sqrt{7}-2}{3}$
• D. $\frac{4-\sqrt{5}}{3}$

Answer 1

The correct option is A $\frac{4-\sqrt{7}}{3}$

$f\left(0\right)=11$, $f\left(1\right)=16$
$f^{\prime }\left(x\right)=3x^2-8x+8$
$\therefore f^{\prime }\left(c\right)=\frac{f\left(1\right)-f\left(0\right)}{1-0}=\frac{16-11}{1}$
$\Rightarrow 3c^2-8c+8=5$
$\Rightarrow 3c^2-8c+3=0$
$\Rightarrow c=\frac{8\pm 2\sqrt{7}}{6}=\frac{4\pm \sqrt{7}}{3}$
As $c\in (0,1)$, so $c=\frac{4-\sqrt{7}}{3}$