question_answer

The minimum value of \[f(x)={{x}^{\frac{2}{3}}}+{{x}^{\frac{1}{3}}},x\in R\]is

A) \[\frac{-1}{2}\]         

B)     \[\frac{-1}{4}\]

C) \[\frac{-1}{8}\]  

D) \[\frac{-1}{16}\]

Correct Answer: B

Solution :

                \[\frac{dy}{dx}=\frac{2}{3}{{x}^{-1/3}}+\frac{1}{3}{{x}^{-2/3}}=\frac{\left( 2{{x}^{1/3}}+1 \right)}{3{{x}^{2}}3}\]                 \[\frac{dy}{dx}=0;\,{{x}^{1/3}}=\frac{-1}{2}\,so\,x=\frac{-1}{8}\] \[f\left( \frac{-1}{8} \right)=\frac{-1}{4}\]