The maximum value

Solution:

(a) $e^{\frac{1}{e}}$

Given : $f(x)=x^{\frac{1}{x}}$

Taking log on both sides, we get

$\log f(x)=\frac{1}{x} \log x$

Differentiating w.r.t. $x$, we get

$\frac{1}{f(x)} f^{\prime}(x)=\frac{-1}{x^{2}} \log x+\frac{1}{x^{2}}$

$\Rightarrow f^{\prime}(x)=f(x) \frac{1}{x^{2}}(1-\log x)$

$\Rightarrow f^{\prime}(x)=x^{\frac{1}{x}}\left(\frac{1}{x^{2}}-\frac{1}{x^{2}} \log x\right)$                      .......(1)

$\Rightarrow f^{\prime}(x)=x^{\frac{1}{x}-2}(1-\log x)$ 

For a local maxima or a local minima, we must have

$f^{\prime}(x)=0$

$\Rightarrow x^{\frac{1}{x}-2}(1-\log x)=0$

$\Rightarrow \log x=1$

$\Rightarrow x=e$

Now,

$f^{\prime \prime}(x)=x^{\frac{1}{x}}\left(\frac{1}{x^{2}}-\frac{1}{x^{2}} \log x\right)^{2}+x^{\frac{1}{x}}\left(\frac{-2}{x^{3}}+\frac{2}{x^{3}} \log x-\frac{1}{x^{3}}\right)=x^{\frac{1}{x}}\left(\frac{1}{x^{2}}-\frac{1}{x^{2}} \log x\right)^{2}+x^{\frac{1}{x}}\left(-\frac{3}{x^{3}}+\frac{2}{x^{3}} \log x\right)$

At $x=e:$

$f^{\prime \prime}(e)=e^{\frac{1}{e}}\left(\frac{1}{e^{2}}-\frac{1}{e^{2}} \log e\right)^{2}+e^{\frac{1}{e}}\left(-\frac{3}{e^{3}}+\frac{2}{e^{3}} \log e\right)=-e^{\frac{1}{e}}\left(\frac{1}{e^{3}}\right)<0$

So, $x=e$ is a point of local maxima.

$\therefore$ Maximum value $=f(e)=e^{\frac{1}{e}}$

Disclaimer: The answer given in the book is incorrect. The solution provided here is according to the question.