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# am i doing this right

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If $h(x)$ is a function whose domain is $[-8,8]$, and $g(x)=h\left(\frac x2\right)$, then the domain of $g(x)$ is an interval of what width?

So I set $h(x) = \sqrt{64-x^2}$ then $g(x) = \sqrt{64-(x/2)^2}$

Here's a graph https://www.desmos.com/calculator/pmgjsaz6jh

So the width of the interval is 32 right?

Feb 3, 2021

#1
+112513
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I got 32 as well but I do not understand your logic.

I just said that for the g function

$$-8\le\frac{x}{2}\le8\\ so\\ -16\le x\le16\\$$

width =32

Feb 3, 2021
#2
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I just set $h(x)$ as a function that fits the question

Guest Feb 3, 2021
#3
+112513
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ok, that is a bit weird but at least I understand what you did now.  Thanks :)

Melody  Feb 3, 2021
#4
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It was a hard question so I wanted to make sure it wasn't a trick question

Guest Feb 3, 2021