# Given f(x)=x^3+2 and g(x)=root3(x-2) how do you show they are inverses?

Jul 19, 2016

It's not elegant, and I'm sure someone will come along soon with a more-elegant demonstration, but one way would be to substitute in, say, -2, 0 and 2 for x in f(x), find the solution, then substitute that solution into g(x) and see whether it yields the original number. Also the reverse.

#### Explanation:

Completing f(x) then g(x):

For -2:

$f \left(- 2\right) = {\left(- 2\right)}^{3} + 2 = - 6$
$g \left(- 6\right) = \sqrt[3]{- 6 - 2} = - 2$

For 0:

$f \left(0\right) = {0}^{3} + 2 = 2$
$g \left(2\right) = \sqrt[3]{2 - 2} = 0$

For 2:

$f \left(2\right) = {2}^{3} + 2 = 10$
$g \left(10\right) = \sqrt[3]{10 - 2} = 2$

Completing g(x) then f(x):

For -2:

$g \left(- 2\right) = \sqrt[3]{- 2 - 2} = - 1.5874$
$f \left(- 1.5874\right) = {\left(- 1.5874\right)}^{3} + 2 = - 2$

For 0:

$g \left(0\right) = \sqrt[3]{0 - 2} = - 1.2599$
$f \left(- 1.2599\right) = {\left(- 1.2599\right)}^{3} + 2 = 0$

For 2:

$g \left(2\right) = \sqrt[3]{2 - 2} = 0$
$f \left(0\right) = {0}^{3} + 2 = 2$

Jul 20, 2016

See explanation. Short answer: switch $x$ and $y$ and solve for $y$.

#### Explanation:

To find the inverse of a function, simply switch the $x$ and $y$ variables and solve for $y$.

Let's apply this concept to the first function given and see if we end up with the second function. Also, note $f \left(x\right)$ is basically $y$.

$f \left(x\right) = {x}^{3} + 2$

$y = {x}^{3} + 2$

First switch the variables.

$x = {y}^{3} + 2$

Now we need to solve for $y$, so isolate $y$.

$x - 2 = {y}^{3} \cancel{+ 2 - 2}$

$x - 2 = {y}^{3}$

$\sqrt[3]{x - 2} = \sqrt[3]{{y}^{3}}$

$\sqrt[3]{x - 2} = y$

We have solved for $y$. Now let's set it equal to the second function and determine if they match.

$\sqrt[3]{x - 2} = \sqrt[3]{x - 2}$

The two functions are identical. Therefore, they are inverses.

Jul 20, 2016

See explanation...

#### Explanation:

Note that the Real cube root is by definition the inverse of the Real cube function. That is: $\sqrt[3]{{t}^{3}} = t$ and ${\left(\sqrt[3]{t}\right)}^{3} = t$ for any Real value of $t$.

So given:

$f \left(x\right) = {x}^{3} + 2$

$g \left(x\right) = \sqrt[3]{x - 2}$

Then:

$f \left(g \left(x\right)\right) = {\left(\sqrt[3]{x - 2}\right)}^{3} + 2 = \left(x - 2\right) + 2 = x$

$g \left(f \left(x\right)\right) = \sqrt[3]{\left({x}^{3} + 2\right) - 2} = \sqrt[3]{{x}^{3}} = x$

So $f$ and $g$ are mutual inverses of one another.

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Footnote

The answer above is about Real functions of Real values. The same does not hold if $f$ and $g$ are considered as Complex valued functions of Complex values, essentially because $t \to {t}^{3}$ is not one to one as a Complex function.

For example ${\left(- \frac{1}{2} + \frac{\sqrt{3}}{2} i\right)}^{3} = 1$, so we have:

$g \left(f \left(- \frac{1}{2} + \frac{\sqrt{3}}{2} i\right)\right)$

$= g \left({\left(- \frac{1}{2} + \frac{\sqrt{3}}{2} i\right)}^{3} + 2\right)$

$= g \left(1 + 2\right)$

$= g \left(3\right)$

$= \sqrt[3]{3 - 2}$

$= \sqrt[3]{1} = 1$