# How do you graph the function f(x)=x^3-2 and its inverse?

Jul 4, 2017

See explanation

#### Explanation:

To graph $f \left(x\right) = {x}^{3} - 2$ it is helpful to remember what the graph of $f \left(x\right) = {x}^{3}$ looks like (See below):

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

graph{x^3 [-10, 10, -5, 5]}

To graph $f \left(x\right) = {x}^{3} - 2$, all we really doing is shifting the $x$-values down $2$ units.

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

graph{x^3-2 [-10, 10, -5, 5]}

As for graphing we first must find the inverse, ${f}^{- 1} \left(x\right)$. We do this by switching the roles of $x$ and $y$ and solving for y.

We can rewrite $f \left(x\right) = {x}^{3} - 2$ as $y = {x}^{3} - 2$ which then becomes $x = {y}^{3} - 2$ when we switch the roles of $x$ and $y$

Now, we solve for $y$

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

x+2=y^3+cancel(-2+2#

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

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

Our inverse function can then be rewritten as:

${f}^{- 1} \left(x\right) = \sqrt[3]{x + 2}$

To graph ${f}^{- 1} \left(x\right) = \sqrt[3]{x + 2}$, take the function of $f \left(x\right) = \sqrt[3]{x}$ and shift the $x$-values $2$ units to the left:

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

graph{root(3)(x+2) [-10, 10, -5, 5]}

The graph of the function is below along with the original function:

graph{(y-x^3+2)(y-root(3)(x+2))=0 [-10, 10, -5, 5]}