For the graph $y=x^{-1/x}$, why are there no real solutions when $x$ is negative?

Question

I have plotted a graph yet it shows no real solutions when $x$ is negative, however I have tried certain values (such as $x=-0.5$) where there is a solution. Help please!

Answer 1

What is the domain of $x^{\frac 1x}$ ?

If you define $\mathbb Q_{odd}=\{\frac pq\mid p\in\mathbb Z,q\in\mathbb N\text{ and }q\text{ odd and }\gcd(p,q)=1\}$

Then $x^y$ is defined for $y\in\mathbb Q_{odd}$ when $x<0$ due to the parity nature (i.e. odd) of $x\mapsto x^n$ function when $n$ is odd (so its inverse is well defined).

Applying this to $x^{-\frac 1x}$ when need $-\frac 1x\in\mathbb Q_{odd}$ in particular $x$ needs to be a rationnal $x=\frac ab$ thus $-\frac 1x=-\frac ba$ and we need $a$ odd.

So on the negative axis your function will also be defined for negative irreductible fractions with odd numerator.

Graphical tools should in theory show a dotted graph but they usually remain silent and show nothing because it is too complicated to handle such a drawing.

Answer 2

Take $x=-\pi$ then $$ y = (-\pi)^{-\frac{1}{-\pi}} = (\pi e^{i\pi})^{\frac{1}{\pi}} = \pi^\pi e^{i} \in \mathbb C $$ Take $x=-2$ then $$ y = (-2)^{-\frac{1}{-2}} = (-2)^{\frac{1}{2}} = \pm i\sqrt{2} \in \mathbb C $$ they are not in $\mathbb R$ but in $\mathbb C$ since

Answer 3

HINT:

Even roots of negative values aren't real. Example: $\sqrt{-2}$ ,i.e. when $x=-2$