$Let\space x\in\mathbb{R}^{-}\Rightarrow x=|x|e^{in\pi}\space where\space n\in\{x:x\space is\space odd\space integer\}$ $\Rightarrow x^x = |x|^xe^{in\pi x}=|x|^x(\cos(xn\pi) + i\sin(xn \pi) )$ $\because x^x\in\mathbb{R} \therefore sin(xn\pi)=0$ $\Rightarrow xn\in\mathbb{N}$ $\Rightarrow x=\dfrac{m}{n}\space where\space n\in\{n:n\space is\space odd\space integer,n\cancel{=}0\},m\in\mathbb{N}$ $\boxed{x^x\in\mathbb{R}\iff \exists p,q\in\mathbb{Z}-\{0\}:\left(x=\dfrac{p}{q}\right)\wedge \left(\dfrac{q}{2}\cancel{\in}\mathbb{Z}\right)}$ $now\space consider\space some\space odd \space integer \space n:0>x=|x|e^{in\pi}$ $x^x=e^{x\ln x}=e^{x(in\pi+\ln|x|)}=e^{xin\pi}\times e^{x\ln|x|}=(-1)^x|x|^x\in\{|x|^x,-|x|^x\}$ $\Rightarrow x^x\in\mathbb{R}\iff x^x\in \{|x|^x,-|x|^x\}$ $\Rightarrow\forall x<0, x\geq -|x|^x=-e^{x\ln|x|}$ $\blue{-\dfrac{de^{x\ln|x|}}{x}=0\Rightarrow -|x|^x(1+\ln|x|)=0}$ $\blue{\because |x|^x\cancel{=}0\Rightarrow \ln|x|=-1\Rightarrow |x|=e^{-1}}\blue{\Rightarrow x=-e^{-1}}\blue{(\because x<0)}$ $\Rightarrow \boxed{\forall x<0,x^x\in\mathbb{R}\Rightarrow x^x>-|-e^{-1}|^{-e^{-1}}=-e^{e^{-1}}=-1.44466786}$

There is already a good solution posted here. I just want to provide an explanation of the "Considering" step for those that may need it.

The reason we can assert that $(-1)^x=-1$ is that such $x$ values densely populate the real number line, which may first seem strange if the way you came to know Euler's formula, $e^{ix}=\cos(x)+i\sin(x)$ , is through alone the visual intuition of tracing a circle.

Let's consider an alternative way of evaluating $(-1)^x$ when rational $x=\frac ab$ ; $(-1)^x=((-1)^a)^{\frac1b}$ . Through this perspective, it becomes clear that if $a$ is even, then $(-1)^x$ is any $b$ th root of unity. Similarly, if $a$ is odd, then $(-1)^x$ is any $b$ th root of $-1$ .

Because $-1$ raised to an odd number equals $-1$ , $-1$ is always one of the values of $(-1)^x$ when both $a$ and $b$ are odd.

So when we assert that $(-1)^x=-1$ , we're assuming that $x$ is a fraction whose numerator and denominator are both odd. And, again, this isn't an unreasonable assertion, because such fractions densely populate the real number line.

let lnf(x)=xln|x|

so f`(x)/f(x)=ln|x|+1

we got that f`(x)=(ln|x|+1) ·f(x)

then we let f`(x)=0 to find the extremum

we got that x= 1/e or -1/e

obviously,f(1/e )>f-1/e)

so the answer is f(-1/e) =-1.44467

New answer:

Obviously,if we want to find the infimum of f(x),we should talk about the situation when x < 0;

make t = -x,I think "let f(x) = x^x be a real function of real variable x" may means we should talk about $g(t)=- t^{-t}$ ,t>0

$g(t)=-e^{-tlnt}$ so $g'(t)=(lnt+1) \cdot e^{-tlnt}$

let f`(x)=0 to find the extremum,we got that $t_0=\frac 1e$

and $g(t_0)=-1.44467$