If \(f(x)=3 \sqrt[3]{x^2}-x^2\), then

• A

  f has no extrema
• Correct answer
  B

  \(f\) is maximum at two points \(x=1\) and \(x=-1\)
• Correct answer
  C

  \(f\) is minimum at \(x=0\)
• D

  f has maximum at \(x=1\) only

Solution

Hint : \(f(x)=3(x)^{\frac{2}{3}}-x^2\) Differentiate both sides wrt. \(x\) \(=f^{\prime}(x)=3 \cdot \frac{2}{3} x^{-1 / 3}-2 x=2 x^{-1 / 3}-2 x=2\left(x^{-1 / 3}-x\right)=2\left(\frac{1-x^{4 / 3}}{x^{1 / 3}}\right)\)
\(\therefore f\) has maximum at two points \(x=-1 x=1\) \(f\) has minimum at \(x=0\)

Hence, the correct answer is options: 'B', 'C'.