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I am unclear if I should consider the function's domain before or after raising it to the power. My textbook gives the following definition of raising a function to a power:

By $f^n$, we mean the function that assigns to $x$ the value $[f(x)]^n$.

Daniel Parker
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2 Answers2

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Note that $h(x) = f^4(x)$ is the composition of $2$ functions: $(g \circ f)(x)$, with $g(x) = x^4$. Thus the domain of $h$ must be $D_{h} = \{x \colon x \in D_{f} \space \text{and}\space f(x) \in D_{g}\} = [0,\infty)$

DeepSea
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    This is the correct way of handling this. However it looses some of its elegance by the way $D_h$ is defined. $D_h=D_{g\circ f}=D_f$ would have been enough. It is the existence of the composition that allows you to say something about its domain. And the existence of it allready ensures that $f(x)\in D_h$ is true for every $x\in D_f$. – drhab Oct 17 '14 at 06:47
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    @drhab With this argumentation, the domain should be given in the first place, but we are looking for the "natural" domain. What if we had $h(x)=\frac1{1-f(x)^4}$ instead? One would argue excatly like OC-Sansoo did to show that $D_h=[0,\infty)\setminus{1}$ – Hagen von Eitzen Oct 17 '14 at 06:47
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    @drhab The composition can indeed exist. It just may not be defined at every point that $f$ was. Take $f(x)=\sqrt{x}$ and $g(x)=\frac1x$. Does $\frac1{\sqrt{x}}$ "not exist"? $0$ is in the domain of $f$ but not in the domain of $g\circ f$. – 2'5 9'2 Oct 17 '14 at 06:48
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It is Still $[0,+\infty)$, since the domain of $f(x)$ is $[0,+\infty)$.

Note that there exists difference between that $f^4(x)$ and the function $x^2$.

Paul
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