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Question

Mathematics Question on Differential Calculus

Let f(x) = x2 logx, x > 0. Then the minimum value of f is

A

1e\frac{1}{\sqrt{e}}

B

2e

C

-2e

D

√e

E

12e\frac{-1}{2e}

Answer

12e\frac{-1}{2e}

Explanation

Solution

The correct option is (E): 12e\frac{-1}{2e}