WBJEE · Maths · Application of Derivatives
If \(f(x)=3 \sqrt[3]{x^2}-x^2\), then
- A f has no extrema
- B \(f\) is maximum at two points \(x=1\) and \(x=-1\)
- C \(f\) is minimum at \(x=0\)
- D f has maximum at \(x=1\) only
Answer & Solution
Correct Answer
(C) \(f\) is minimum at \(x=0\)
Step-by-step Solution
Detailed explanation
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 \(x=-1 x=1\) \(f\)…
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