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The function $f(x)=x^x$ decreases on the interval |
$(0, e)$ $(0, 1)$ $(0,\frac{1}{e})$ None of these |
$(0,\frac{1}{e})$ |
Clearly, f(x) is defined for x > 0 Now, $f(x)=x^x⇒f'(x)=x^x(1+\log x)$ For to be decreasing, we must have $f'(x)<0⇒x^x(1+\log x)<0$ $⇒1+\log x<0⇒\log x< -1 ⇒ x<e^{-1}$ So, f(x) is decreasing on $(0,\frac{1}{e})$ |
