My coaching module says "for domain of $ k(x)= f(x) ^ {g(x)} $ conventionally the conditions are $f(x)>0$ and $g(x)$ must be real." I get why $g(x)$ must be real, but why must $f(x)>0$ ? If for instance $f(x) = -1$,why wouldn't $k(x)$ be real? Please help me in understanding this.
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7What is $(-1)^\pi$? – José Carlos Santos Sep 10 '18 at 12:45
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The expression $x^p$ for $x\le0$ is well defined only for some $p=\frac{m}n$ rational with $n$ odd therefore for the real functions in the form $f(x)^{g(x)}$ we require that $f(x)>0$ even if also for $f(x)\le 0$ for some value of $g(x)$ the expression can be well defined.
For example
$(-1)^{\frac12}=N.D.$
$(-1)^{\frac23}=1$
- $0^{\frac23}=0$
- $0^{-\frac23}=N.D.$
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