You can find the inverse of a relation or the inverse of a function. If you are given a set of (x, y) coordinate points, you reverse the domain (x) with the range (y) to create the inverse. Relations are inverses of each other when the domain of the function becomes the range of the inverse.
For Example
Find the inverse of the relation
{(-3, -20); (-1, -12); (0, -8); (1, -4); (3, 4)}
Since this relation is given as (x, y) coordinate pairs, simply reverse the order of each pair to create the inverse. |
The inverse is {(-20, -3); (-12, -1); (-8, 0); (-4, 1); (4, 3)} |
If you need to find the inverse of a function, change the notation from \(f(x)\) to y, then interchange the x and the y in the equation. Resolve the equation for y. For example:
Find the inverse of the function \( f\left( x \right) = 7x^{2} - 16 \).
Change the notation from \(f(x)\) to y. |
\( f\left( x \right) = 7x^{2} - 16 \) \( y = 7x^{2} - 16 \) |
Interchange the x and the y in the equation. |
\( x = 7y^{2} - 16 \) |
Resolve the equation for y. Note that the inverse is not a function since each input will have more than one output. |
\( 7y^2 = x + 16 \) \( y^2 = \frac{x+16}{7} \) \( y^{2} \) = \( \pm \sqrt{\frac{x+16}{7}} \) y = \( \pm \sqrt{\frac{x+16}{7}} \) |
Think you got it?
Use the activity below to practice finding the inverses of relations and functions. Find the inverse of the relation or function that is shown, then check your answer.
What is the inverse of \( f\left( x \right) = x^{3} + 3 \)?
\( f^{- 1}\left( x \right) = \sqrt[3]{x - 3} \)
If you need help arriving at this answer, click the solution button.
First, rewrite f(x) to y. |
\( y = x^3 + 3\) |
Then, interchange the y and x variables. |
\( x = y^3 + 3\) |
Next, solve for y. |
\( x - 3 = y^3 \) The inverse function is \(f^{-1}(x) \) = \( \sqrt[3]{x - 3} \). |
What is the inverse of \( f\left( x \right) = x^{2} - 1 \)?
\( f^{- 1}\left( x \right) = \pm \sqrt{x + 1} \)
If you need help arriving at this answer, click the solution button.
First, rewrite f(x) to y. |
\(y = x ^2 - 1\) |
Then, interchange the y and x variables. |
\( x = y^2 - 1 \) |
Next, solve for y. |
\(x + 1 = y ^2 \) |
