The function


The function $f(x)=x^{x}$ has a stationary point at

(a) $x=e$

(b) $x=\frac{1}{e}$

(c) $x=1$

(d) $x=\sqrt{e}$


The values of $x$ for which $f^{\prime}(x)=0$ are called stationary values.

Let $f(x)=x^{x}$.


$\Rightarrow \log f(x)=\log x^{x}$

$\Rightarrow \log f(x)=x \log x$

Differentiating both sides with respect to x, we get

$\frac{1}{f(x)} \times f^{\prime}(x)=x \times \frac{1}{x}+\log x \times 1$

$\Rightarrow f^{\prime}(x)=x^{x}(1+\log x)$

For stationary point, we have


$\Rightarrow x^{x}(1+\log x)=0$

$\Rightarrow 1+\log x=0 \quad\left[x^{x}>0\right]$

$\Rightarrow \log x=-1$

$\Rightarrow x=e^{-1}=\frac{1}{e}$

Thus, the function $f(x)=x^{x}$ has a stationary point at $x=\frac{1}{e}$.

Hence, the correct answer is option (b).

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