Question #40d14

May 18, 2017

${f}^{-} 1 \left(x\right) = \sqrt[3]{x - 1} + 3$

Explanation:

You are asking to find the inverse function of $f \left(x\right)$ or ${f}^{-} 1 \left(x\right)$

When finding the inverse of a function, you are essentially solving for $x$ but we could also simply switch the $x$ and $y$ variables in the equation above and solve for $y$ like any other problem such that:

Let $f \left(x\right) = y$ so...
$y = {\left(x - 3\right)}^{3} + 1 \to x = {\left(y - 3\right)}^{3} + 1$

Now we can solve for $y$

Subtract $1$ from both sides:

$x - 1 = {\left(y - 3\right)}^{3} + \cancel{1 - 1}$

$x - 1 = {\left(y - 3\right)}^{3}$

Take the cubed root on each side to cancel out the exponent of $3$

$\sqrt[3]{x - 1} = \sqrt[3]{{\left(y - 3\right)}^{3}}$

$\sqrt[3]{x - 1} = y - 3$

Finally, add $3$ to both sides

$\sqrt[3]{x - 1} + 3 = y \cancel{- 3 + 3}$

$\sqrt[3]{x - 1} + 3 = y$ (This is our inverse function)

I mentioned earlier that finding the inverse of a function means that you are solving for $x$ but I also suggested that you could simply just switch $x$ and $y$ and solve for $y$ instead. What I'm going to do now is to show the solution in which we solve for $x$ instead of $y$. You'll find that the process is exactly the same with a little tweak at the end:

$y = {\left(x - 3\right)}^{3} + 1$

Subtract $1$ from both sides:

$y - 1 = {\left(x - 3\right)}^{3} + \cancel{1 - 1}$

$y - 1 = {\left(x - 3\right)}^{3}$

Take the cubed root on each side to cancel out the exponent of $3$

$\sqrt[3]{y - 1} = \sqrt[3]{{\left(x - 3\right)}^{3}}$

$\sqrt[3]{y - 1} = x - 3$

Finally, add $3$ to both sides

$\sqrt[3]{y - 1} + 3 = x \cancel{- 3 + 3}$

$\sqrt[3]{y - 1} + 3 = x$

As you can see, the equation above is almost exactly the same as the other one we solved for except this function is written in terms of $x$. The tweak I was talking about is that you could choose to solve for $x$ from the very beginning but you must remember to switch the variables $x$ and $y$ at the end so that your answer is expressed in terms of $y$. Thus,

$\sqrt[3]{y - 1} + 3 = x \to \sqrt[3]{x - 1} + 3 = y$ (This is our inverse function)

We can then let $y$ be ${f}^{-} 1 \left(x\right)$, the inverse of $f \left(x\right)$

$\therefore {f}^{-} 1 \left(x\right) = \sqrt[3]{x - 1} + 3$