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I'm trying to find the domain and the form of the composition of the following functions: $f:y=\sqrt[3]{x}$ , $g:y=1-x^2$.

My solution is the following: domain = $[-1,1]$ , form = $f\circ g: y= 1-x^3$

My steps:

domain: $D(f\circ g)=${$x\in\mathbb{R}, 1-x^2\in\mathbb{R}^+$}, $\Rightarrow 1-x^2\geq 0$ $\Leftrightarrow x^2\leq 1$ $\Rightarrow x\in[-1,1]$

form: $1-(\sqrt{x^3})^2 \Leftrightarrow 1-x^3$

Is it correct? Thanks!

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  • $\begingroup$ Why do you end up with those? $\endgroup$ Commented Oct 5, 2019 at 15:23
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    $\begingroup$ I think, you should get $\sqrt[3]{1-x^2}$. $\endgroup$ Commented Oct 5, 2019 at 15:25
  • $\begingroup$ And the domain is $\mathbb{R}$ $\endgroup$ Commented Oct 5, 2019 at 15:25
  • $\begingroup$ @Gae.S. hmm.. I will edit the post with my steps. $\endgroup$ Commented Oct 5, 2019 at 15:26
  • $\begingroup$ @HVxvejjw Updated the post with my steps, could you please tell me where I did a mistake? $\endgroup$ Commented Oct 5, 2019 at 15:32

1 Answer 1

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To find the domain of the composite function

$$( f \circ g )(x)=f(g(x))$$

one can follow these two steps

  1. Find the domain of the inside (input) function.
  2. Construct the composite function. Find the domain of this new function. If there are restrictions on this domain, add them to the restrictions from Step 1. If there is an overlap, use the more restrictive domain (or the intersection of the domains).

In your case, the domain of the inside function $g(x)=1-x^2$ is all real numbers while the domain of $f(x)=\sqrt[3]{x}$ is also all real numbers.

The domain of the composite function is where $f(g(x))=f(1-x^2)=\sqrt[3]{1-x^2}$ is defined. Although, observe that $\sqrt[3]{1-x^2}$ is also defined for all real numbers.

Therefore, the domain of $f(g(x))$ is all real numbers.

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  • $\begingroup$ I edited the post with my steps, could you tell me what I did wrong? $\endgroup$ Commented Oct 5, 2019 at 15:36
  • $\begingroup$ I'd say the domain of $\sqrt[3]\bullet$ is not $[0,\infty)$, but $\Bbb R$. $\endgroup$ Commented Oct 5, 2019 at 15:42
  • $\begingroup$ @Gae.S. I'm really confused right now. Is the cube root defined for negative numbers in $\mathbb{R}$ or not? $\endgroup$ Commented Oct 5, 2019 at 15:48
  • $\begingroup$ Ah! I made the same mistake you did! I shouldn't have trusted this result without drawing the function myself. The domain of the cube root of x is all real numbers. $\endgroup$ Commented Oct 5, 2019 at 15:50
  • $\begingroup$ I don't see a problem with $\sqrt[3]x:=\text{the one and only real number }u\text{ such that }[u^3=x]$, for $x<0$. $\endgroup$ Commented Oct 5, 2019 at 15:50

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