Let f(x)=x−x. Taking logarithm: lnf(x)=−xlnx. Differentiating: f(x)f′(x)=−(1+lnx). Setting f′(x)=0: 1+lnx=0⇒x=1/e. Maximum value =(1/e)−1/e=e1/e.
The maximum value of x^-x is
CUET UG 2022 — Mathematics Calculus
The maximum value of x−x is
Held on 10 Aug 2022 · Verified 13 Jul 2026.
Options
- A
ee−1
- B
ee1
- C
ee−1
- D
ee−4
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