# Finding an Inverse function with multiple occurences of $y$

For some reason I cannot figure out how the book is finding the solution to this problem.

find the inverse function of $f(x)=3x\sqrt{x}$.

My steps seem to lead to a dead end:

step 1. switch $f(x)$ with $y$: $y = 3x\sqrt{x}$

step 2. swap $x$ and $y$: $x = 3y\sqrt{y}$

step 3. solve for $y$:

step 3.1: $\displaystyle \frac{x}{3y} = \sqrt{y}$.

step 3.2: $\displaystyle \left(\frac{x}{3y}\right)^2 = \left(\sqrt{y}\right)^2$

step 3.3: $\displaystyle \frac{x^2}{9y^2} = y$ ... uhhh?

The book answer is $\displaystyle y = \left(\frac{x}{3}\right)^{2/3}$.

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As always, thanks Arturo Magidin for cleaning up the post, and converting the mathematics into LaTeX. – Eric Naslund Feb 14 '11 at 0:54
+1 for showing your work. – Arturo Magidin Feb 14 '11 at 0:54

Nothing wrong with what you've done so far, you just haven't finished!

From $\displaystyle \frac{x^2}{9y^2} = y$, move all the $y$'s to the same side:

step 3.4: $x^2 = 9y^3$.

Now isolate $y$:

step 3.5: $\displaystyle\frac{x^2}{9} = y^3$.

Now solve for $y$ by taking cubic roots; you might also notice that the left hand side is a perfect square...

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Note that $\frac{x^2}{9}=\left(\frac{x}{3}\right)^2$. Also note that you can multiply both sides of your equation by $y^2$ to get just one $y$ term, and that the cube root is the same as the $\frac{1}{3}$ power.

You could, as lhf indicates, avoid ever breaking up the $y$ terms in the first place. Technically this would be preferred, because when you divided by $y$ you made the assumption that $y\neq 0$, which is not always true.

Beware of squaring when solving equations in general, because it can lead to extraneous solutions. In this case it is a valid step, but for a silly example consider trying to solve the equation $\sqrt{y}=-1$ by squaring both sides.

A final remark. The original function has an implicit domain, $x\geq 0$. Your inverse function has a formula definition that does not appear to require any domain restriction, but it is misleading to say that $f^{-1}(x)=\left(\frac{x}{3}\right)^{2/3}$ without mentioning that this is only valid when $x\geq 0$. This domain restriction arises from the fact that the range of $f$ is $[0,\infty)$. The function $h(x)=\left(\frac{x}{3}\right)^{2/3}$ on the whole real line is not invertible (it fails the horizontal line test), whereas if $g(x)=-f(x)=-3x\sqrt{x}$, then the inverse function would be $g^{-1}(x)=\left(\frac{x}{3}\right)^{2/3}$ on $(-\infty,0]$. Looking at the graphs of these functions and keeping in mind how inverses are reflected across the line $y=x$ (resulting from switching $x$ and $y$) should make it clearer what is going on.

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$x=3y\sqrt y$ implies $x^2=9y^3$.

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The algorithm you are applying is correct, and will yield the inverse function. In step 3 you should square both sides. Then we get $$x^2=9y^2\cdot y=9y^3$$

Can you solve the problem from here?

Hint: Divide by $9$ and then take cube roots.

Hope that helps.

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Two hints: $\sqrt{x}=x^{\underline{?}}$ and $x^a\cdot x^b=x^{\underline{?}}$

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