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This question seems quite straightforward, but it isn't. Let $f^{-1}(x)=g(x)$

$f(x)=\left\{\begin{array}{l}x^3-1, x<2\\x^2+3,x\geq2\\\end{array}\right.$

$f(g(x))=x$

If $x<2,(g(x)) ^3=x+1 \implies g(x)=(x+1)^{1/3}, x<2$

Similarly, if $x\geq2, g(x)=(x-3)^{1/2}$. We know that $f(x)$ is one one function, so it passes the horizontal line test.

However my book says $g(x)=\left\{\begin{array}{l}(x+1)^{1/3}, x<7\\(x-3)^{1/2},x\geq7\\\end{array}\right.$

Edit 1: The strange thing is that when I substitute values, the answer given in my books seems correct, even though it doesn't make sense mathematically.

Edit 2: Since we have to consider the domain of another variable $y$ is it appropriate to write the inverse as a function of $x$.

How did this happen?

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    $\begingroup$ Why do you consider the $x<2$ case? Think again. $\endgroup$
    – Trebor
    Sep 11, 2021 at 6:04
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    $\begingroup$ When $x < 2$ and $y = x^3 - 1$ then [1] $y < 7$ and [2] $y + 1 = x^3 \implies (y + 1)^{(1/3)} = x.$ $\endgroup$ Sep 11, 2021 at 6:10
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    $\begingroup$ The book's answer would probably have been clearer if it had said that $f(x) = y, g(y) = x$, and that $g(y) = (y + 1)^{(1/3)} ~: ~y < 7.$ $\endgroup$ Sep 11, 2021 at 6:13

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Infinity_hunter gave you the correct solution so let me tell you your mistake. Where you considered $x<2$ is not correct. You should consider $g(x)<2$ as the restriction is in the input value of f then everything will be clear.

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Hint:

Consider the case $x<2$. Then $y = f(x) = x^3-1$ and let $g(y)$ denote the inverse. Since $x<2$ we see that $y < 2^3-1=7$. So $x = (y+1)^{\frac13} = g(y)$ where $y <7$. So when $y<7$ we have $g(y) = (y+1)^{\frac13}$.

I hope that you can do the other part as well.

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