+1 vote
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The function $f(x) = x^{1/x}, \: x \neq 0$ has

1. a minimum at $x=e$;
2. a maximum at $x=e$;
3. neither a maximum nor a minimum at $x=e$;
4. None of the above
in Calculus
edited | 54 views

Solution:

Let $y = f(x) = x^{1/x}$

Taking $\log$ on both sides, we get:

$\log y = \log(x^{1/x}) \implies \log y = \frac{1}{x}\log x$

Differentiating above w.r.t. x, we get:

$\frac{1}{y}\frac{dy}{dx} = \frac{1}{x^2} - \frac{1}{x^2}\log x = 0$

Now,

$\frac{dy}{dx} = y \underbrace{ \Bigg(\frac{1}{x^2} - \frac{1}{x^2}\log x \Bigg)}_{\text = 0}= 0 \implies \frac{dy}{dx} = y\frac{1}{x^2}(1-\log x)$

Here, $\frac{1}{x^2}$ can't be zero.

So,$1-\log x = 0 \implies \log x = 1 \implies e ^1 = x \implies x = e$

$\therefore$ The function has a maximum at $x = e$

Hence,  B is the correct option.

by Boss (18.9k points)
edited by
0
But you haven't proved whether the function has the maximum or the minimum at $x=e$.

For this, you have to take $2^{\mathrm{nd}}$ derivative as well. By the way, the function has the maximum at $x=e$, because $f''(e)<0$.
0
It won't be $f^{''}(e)$, but $f^{''}(x)$ is just fine.
0
I skipped that because I felt that it was too obvious. Also that is theoretical exam point of view not needed here, since I already got the value.