Question

The value of $c$ in the lagranges mean value theorem for $f\left(x\right)=x^3,a=1,h=\frac{1}{2}$ is

• A. $\frac{1}{3}$
• B. $\sqrt{\frac{19}{56}}$
• C. $\sqrt{\frac{19}{3}}+2$
• D. $\sqrt{\frac{19}{3}}-2$

Answer 1

The correct option is D $\sqrt{\frac{19}{3}}-2$

$f\left(x\right)$ is continous & differentiable

$f^{\prime }\left(x\right)=3x^2$
$f^{\prime }\left(c\right)=\frac{f\left(\frac{3}{2}\right)-f\left(1\right)}{\frac{1}{2}}$
$3c^2=\frac{\frac{27}{8}-1}{\frac{1}{2}}$
$3c^2=\frac{19}{4}$

$c=\frac{1}{2}\sqrt{\frac{19}{3}}$

$\therefore cϵ(1,\frac{3}{2})$