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All the graph plotting tools do not display values for positive $x$ - why? It seems valid for me, for example when $x = 2$, $y = (-2)^2 = 4$

Pedro
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Draif Kroneg
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In programming languages $a^b$ is usually constructed as $a^b := e^{b\log a}$, and for the real case $\log x$ is only defined for strictly positive $x$.

flawr
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    I just explained why those programs behave the way they do. This is not about what is well defined. – flawr May 28 '17 at 20:44
  • A rational number with an even numerator (such as 2/3) it is positive and for a rational number with odd numerator and denominator (such as 2000001/3000001) it is negative. for a rational with an even denominator (such as 2000001/3000000) it is undefined and for any irrational (such as $\frac {\sqrt{4.000001}}3$) it is undefined . If you wan't to write a graphing calculator that can handle that,, be my guest. – fleablood May 28 '17 at 20:58
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Hint :

$$\sqrt[2n]{x}=\text{Undefined} : \ \ \ n \in \mathbb{R} , x <0 $$

such as :

$$\sqrt[2]{\dfrac{-1}{2}}=(-\dfrac{1}{2})^{\dfrac12}=\text{Undefined}$$

$$\sqrt[4]{\dfrac{-1}{4}}=(-\dfrac{1}{4})^{\dfrac14}=\text{Undefined}$$

Community
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Almot1960
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By definition if $h(x)=\left(f(x)\right)^{g(x)}$ then a domain of $h$ is $\{x|f(x)>0\}\cap$ domain of $g$.

Why do we do it?

Because we wish to work with function $f(x)=(-x)^{x}$.

For example, $(-1)^1=-1$, but $(-0)^0$ is not defined.

$\left(-\frac{1}{2}\right)^{\frac{1}{2}}$ is not defined, but $\sqrt[3]{-\frac{1}{3}}=-\frac{1}{\sqrt[3]{3}}$.

Id est, sometimes we can calculate a value of $f$ for $x\geq0$ and sometimes we can not do it.

It's not comfortable for a work with this function (derivative,...).

Thus, the domain of $f$ is $(-\infty,0)$, but now by definition.

Michael Rozenberg
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