Question

The critical points of the function $f\left(x\right)=x^{3/5}(4-x),x\in R^{+}\cup \left\{0\right\}$ is _______

• A. $0,-\frac{3}{2}$
• B. $0,\frac{3}{2}$
• C. $0,\frac{2}{3}$
• D. $0,-\frac{2}{3}$

Answer 1

Answer: (C)

$0,\frac{2}{3}$

$f\left(x\right)=x^{\frac{3}{5}}(4-x)x\in R^{+}$
$f\left(x\right)=4x^{\frac{3}{5}}-x^{\frac{8}{5}}$
$\therefore f^{\prime }\left(x\right)=4\left(\frac{3}{5}\right)x^{-\frac{2}{3}}=\frac{8}{5}{x}^{\frac{3}{5}}$
$\therefore f^{\prime }\left(x\right)=\frac{4}{5}(\frac{3}{x^{\frac{2}{3}}}-2x^{\frac{3}{5}})=\frac{4}{5}\left(\frac{3-2x}{x^{\frac{2}{3}}}\right)$
For $x=0;f^{\prime }\left(x\right)$ does not exists and for $x=\frac{2}{3}$ we have $f^{\prime }\left(x\right)=0$
$\therefore$ Critical points are $0$ and $\frac{2}{3}$