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Where x squared log x bottoms out

EasyCalculus~3m

The function f(x)=x2ln⁡xf(x)=x^2\ln x (for x>0x>0) has a single interior minimum. At what value of xx does it occur? (to 3 decimals)

Show hints (2)+
  1. Differentiate x2ln⁡xx^2\ln x with the product rule, then set it to 00.
  2. The equation reduces to 2ln⁡x+1=02\ln x + 1 = 0.

Answer

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0.607 (± 0.005)

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Asked at: Multi-Strategy Quant, Data-Driven Research

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