Question #95d2b

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Question #95d2b

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Answer 1 · 2017-11-29T23:53:47

$f^-1(x)=+-sqrt((x+5)/7)$ (In the explanation, we will not use $f^-1(x)$ to make things simpler.

Explanation:

$g(x)$ is inverse of $f(x)$ if $g(f(x))=x$ . If this is true, this is also true: $f(g(x))=x$ .

Let's call our inverse function $g(x)$
We know that $g(f(x))=x$ and that $f(g(x))=x$ .

Out of the two, we will use $f(g(x))=x$ .
Whatever $g(x)$ is, we know that when it is plugged into $f(x)$ , it will give us $x$ . To make matters simpler, we will call $g(x)$ as $y$ .

Therefore, we know that $7y^2-5=x$ Isolate $y$ .
$7y^2-5=x$
$7y^2=x+5$
$y^2=(x+5)/7$
$y=+-sqrt((x+5)/7)$

We see that when these are plugged into $7x^2-5$ , it gives us $x$ .

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Question #95d2b

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