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Q.
Let exp(x) denote exponential function ex. If $f\left(x\right) = exp \left(x^{\frac{1}{x}}\right), \, x > 0$ then the minimum value of f in the interval [2, 5] is
Given that,
$f(x)=e^{(x)^{\frac{1}{x}}, x>0}$
Taking log on both sides, we get
$ \log +(x)=(x)^{\frac{1}{x}}=g(x)(\text { say }) \ldots(i) $
Here,$ g(x)=x^{\frac{1}{x}}$
$\Rightarrow \log g(x)=\frac{1}{x} \log x$
On differentiating w.r.t. $x$, we get
$\cdot \frac{1}{g(x)} \cdot g^{\prime}(x) =\frac{x \cdot \frac{1}{x}-\log x}{x^{2}} $
$=\left(\frac{1-\log x}{x^{2}}\right)$
$\Rightarrow g^{\prime}(x)=x^{\left(\frac{1}{x}-2\right)}(1-\log \bar{x})$
For maximum or minimum of $g(x)$ put $g^{\prime}(x)=0$
$\Rightarrow x^{\left(\frac{1}{x}-2\right)}(1-\log x)=0$
$\Rightarrow \log x=1=\log e$
$\Rightarrow x=e$
and $\left.g^{\prime \prime}(x)\right|_{x=e} > 0$
So, $g(x)$ is minimum at $x=e$
$\therefore g(x)$ increases in $(0, e)$ and decreases in $(e, \infty)$ it will be minimum at either 2 or 5
$\therefore 2^{\frac{1}{2}} > 5^{\frac{1}{5}} $
$\Rightarrow $ Minimum value of $f(x)=e^{(5)^{\frac{1}{5}}}$