The graph of f is shown at left. It passes the horizontal-line test. Thus it is one-to-one and its inverse is a function. We also proved that f is one-to-one in Example 3. We find a formula for $f^{-1}(x)$.

The graphs of f and $f^{-1}$ are shown at left. The solutions of the inverse function can be found from those of the original function by interchanging the first and second coordinates of each ordered pair.

When we interchange x and y in finding a formula for the inverse of $f(x)=2x-3$, we are in effect reflecting the graph of that function across the line $y=x$. For example, when the coordinates of the y-intercept, (0, −3), of the graph of f are reversed, we get the x-intercept, (−3, 0), of the graph of $f^{-1}$. If we were to graph $f(x)=2x-3$ in wet ink and fold along the line $y=x$, the graph of $f^{-1}(x)=(x+3)/2$ would be formed by the ink transferred from f.

1. The graph of $g(x)=x^3+2$ is shown at left. It passes the horizontal-line test and thus has an inverse that is a function. We also know that $g(x)$ is one-to-one because it is an increasing function over its entire domain.

2. We follow the procedure for finding an inverse.

   $$\begin{array}{lrcl}\mathbf{1.}\mathrm{Replace}g(x)\mathrm{with}y:& y& =& x^3+2\\ \mathbf{2.}\mathrm{Interchange}x\mathrm{and}y:& x& =& y^3+2\\ \mathbf{3.}\mathrm{Solve\; for}y:& x-2& =& y^3\\ & \sqrt[3]{x-2}& =& y\\ \mathbf{4.}\mathrm{Replace}y\mathrm{with}{g}^{-1}(x):& g^{-1}(x)& =& \sqrt[3]{x-2}.\end{array}$$

   We can test a point as a partial check:

   $$\begin{array}{rcl}g(x)& =& x^3+2\\ g(3)& =& 3^3+2=27+2=29.\end{array}$$

   Will $g^{-1}(29)=3$? We have

   $$\begin{array}{rcl}{g}^{-1}(x)& =& \sqrt[3]{x-2}\\ g^{-1}(29)& =& \sqrt[3]{29-2}=\sqrt[3]{27}=3.\end{array}$$

   Since $g(3)=29$ and $g^{-1}(29)=3$, we can be reasonably certain that the formula for $g^{-1}(x)$ is correct.

   x	$g(x)$
   −2	−6
   −1	1
   0	2
   1	3
   2	10

3. To find the graph of the inverse function, we reflect the graph of $g(x)=x^3+2$ across the line $y=x$. This can be done by plotting points.

   

   x	$g^{-1}(x)$
   −6	−2
   1	−1
   2	0
   3	1
   10	2