Lesson 5.7 : Inverses and Radical Functions

New Concept

This lesson explores how to find the inverse of polynomial functions. You'll learn why some functions, like quadratics, require a restricted domain to become one-to-one and how this process often leads to radical functions as inverses.

What’s next

Next, you'll tackle interactive examples for finding inverses. We'll start with invertible polynomials and then see how to handle functions requiring domain restrictions.

Property

Two functions, $f$ and $g$, are inverses of one another if for all $x$ in the domain of $f$ and $g$, $g(f(x)) = f(g(x)) = x$. An important relationship between inverse functions is that they “undo” each other. If $f^{-1}$ is the inverse of a function $f$, then $f$ is the inverse of the function $f^{-1}$. In other words, whatever the function $f$ does to $x$, $f^{-1}$ undoes it—and vice-versa.

Examples

• Show that $f(x) = \frac{x+5}{3}$ and $f^{-1}(x) = 3x - 5$ are inverses. We check that $f^{-1}(f(x)) = 3(\frac{x+5}{3}) - 5 = (x+5) - 5 = x$. And $f(f^{-1}(x)) = \frac{(3x-5)+5}{3} = \frac{3x}{3} = x$.
• Show that $f(x) = \sqrt[3]{x-2}$ and $f^{-1}(x) = x^3 + 2$ are inverses. We check that $f^{-1}(f(x)) = (\sqrt[3]{x-2})^3 + 2 = (x-2) + 2 = x$. And $f(f^{-1}(x)) = \sqrt[3]{(x^3+2)-2} = \sqrt[3]{x^3} = x$.
• Show that $f(x) = \frac{1}{x+4}$ and $f^{-1}(x) = \frac{1}{x} - 4$ are inverses for $x \neq 0, -4$. We check that $f^{-1}(f(x)) = \frac{1}{\frac{1}{x+4}} - 4 = (x+4) - 4 = x$. And $f(f^{-1}(x)) = \frac{1}{(\frac{1}{x}-4)+4} = \frac{1}{\frac{1}{x}} = x$.

Explanation

Think of inverse functions as a "reverse" button. If you apply a function and then immediately apply its inverse to the result, you get back to your original number. This works because their compositions cancel each other out, always resulting in $x$.

Property

To find the inverse of a one-to-one polynomial function, follow these steps:

1. Replace $f(x)$ with $y$.
2. Interchange $x$ and $y$.
3. Solve for $y$, and rename the function $f^{-1}(x)$.

Functions that have inverses are called invertible functions. The notation $f^{-1}(x)$ is used for the inverse function and is not the same as the reciprocal $\frac{1}{f(x)}$.

Examples

• Find the inverse of the one-to-one function $f(x) = 2x^3 + 7$. First, write $y = 2x^3 + 7$. Swap variables to get $x = 2y^3 + 7$. Solve for $y$: $x-7 = 2y^3$, so $y^3 = \frac{x-7}{2}$. Thus, $f^{-1}(x) = \sqrt[3]{\frac{x-7}{2}}$.
• Find the inverse of $f(x) = (x-3)^3$. Write $y = (x-3)^3$. Swap variables to get $x = (y-3)^3$. Solve for $y$: $\sqrt[3]{x} = y-3$. Thus, $f^{-1}(x) = \sqrt[3]{x} + 3$.
• Find the inverse of $f(x) = 5x - 9$. Write $y = 5x - 9$. Swap variables to get $x = 5y - 9$. Solve for $y$: $x+9 = 5y$. Thus, $f^{-1}(x) = \frac{x+9}{5}$.

Explanation

Finding an inverse is like reversing the roles of input and output. By swapping the $x$ and $y$ variables in the equation, you are literally switching the input and output. Solving for the new $y$ gives you the rule for the inverse function.

Property

If a function is not one-to-one, it cannot have an inverse. If we restrict the domain of the function so that it becomes one-to-one, thus creating a new function, this new function will have an inverse. To do this:

1. Restrict the domain, often at the vertex for a quadratic, so the function is one-to-one.
2. Replace $f(x)$ with $y$, interchange $x$ and $y$, and solve for $y$.
3. Ensure the formula for $f^{-1}(x)$ gives outputs that match the restricted domain of the original function.

Examples

• Find the inverse of $f(x) = (x-5)^2$ on the restricted domain $x \geq 5$. Let $y=(x-5)^2$, so $x=(y-5)^2$. Then $\pm\sqrt{x} = y-5$, which gives $y = 5 \pm \sqrt{x}$. Since the original domain was $x \geq 5$, the inverse's range must be $y \geq 5$, so we choose the positive case: $f^{-1}(x) = 5 + \sqrt{x}$.
• Find the inverse of $f(x) = (x-5)^2$ on the restricted domain $x \leq 5$. Following the same steps, we get $y = 5 \pm \sqrt{x}$. Since the original domain was $x \leq 5$, the inverse's range must be $y \leq 5$, so we choose the negative case: $f^{-1}(x) = 5 - \sqrt{x}$.
• Find the inverse of $f(x) = x^2 - 1$ on the restricted domain $x \geq 0$. Let $y=x^2-1$, so $x=y^2-1$. Then $x+1=y^2$, which gives $y=\pm\sqrt{x+1}$. Since the original domain was $x \geq 0$, the inverse's range must be $y \geq 0$, so we choose the positive case: $f^{-1}(x) = \sqrt{x+1}$.

Explanation

A U-shaped parabola fails the horizontal line test. To give it an inverse, we must 'slice' it at its turning point (vertex) and use only one branch. This makes it one-to-one, allowing us to find a proper inverse function for that piece.

Property

To find the inverse of a radical function:

1. Determine the range of the original function. The range of $f$ will be the domain of $f^{-1}$.
2. Replace $f(x)$ with $y$, then solve for $x$ by isolating the radical and raising both sides to the appropriate power.
3. Rename the new function $f^{-1}(x)$ and restrict its domain to match the range of the original function.

Examples

• Find the inverse of $f(x) = \sqrt{x-2}$. The range is $f(x) \geq 0$. Let $y = \sqrt{x-2}$, so $x = \sqrt{y-2}$. Squaring both sides gives $x^2 = y-2$, so $y = x^2+2$. The inverse is $f^{-1}(x) = x^2+2$, with domain restricted to the original range, $x \geq 0$.
• Find the inverse of $f(x) = \sqrt{3x+7}$. The range is $f(x) \geq 0$. Let $y = \sqrt{3x+7}$, so $x = \sqrt{3y+7}$. Squaring gives $x^2 = 3y+7$, so $y = \frac{x^2-7}{3}$. The inverse is $f^{-1}(x) = \frac{x^2-7}{3}$, with domain $x \geq 0$.
• Find the inverse of $f(x) = \sqrt[3]{x+5}$. A cube root has a range of all real numbers. Let $y = \sqrt[3]{x+5}$, so $x = \sqrt[3]{y+5}$. Cubing both sides gives $x^3 = y+5$, so $y=x^3-5$. The inverse is $f^{-1}(x) = x^3-5$, with no domain restriction.

Explanation

To reverse a radical (like a square root), you use its opposite operation (squaring). Since the original radical function has a restricted output (e.g., $\sqrt{x}$ is always non-negative), its inverse must have a restricted input to be a true inverse.

Property

To find the inverse of a rational function, we solve for the independent variable.

1. Replace the function notation with a variable.
2. Multiply both sides by the denominator.
3. Distribute and gather all terms with the variable you are solving for on one side.
4. Factor out the variable.
5. Solve by dividing.

Examples

• Find the inverse of $f(x) = \frac{x+2}{x-5}$. Let $y=\frac{x+2}{x-5}$, so $x=\frac{y+2}{y-5}$. Then $x(y-5)=y+2$, so $xy-5x=y+2$. Next, $xy-y=5x+2$, so $y(x-1)=5x+2$. The inverse is $f^{-1}(x)=\frac{5x+2}{x-1}$.
• Find the inverse of $f(x) = \frac{4x}{x+3}$. Let $y=\frac{4x}{x+3}$, so $x=\frac{4y}{y+3}$. Then $x(y+3)=4y$, so $xy+3x=4y$. Next, $3x=4y-xy$, so $3x=y(4-x)$. The inverse is $f^{-1}(x)=\frac{3x}{4-x}$.
• For a solution with concentration $C = \frac{30+0.2n}{120+n}$, find $n$ in terms of $C$. Start with $C(120+n)=30+0.2n$, so $120C+Cn=30+0.2n$. Then $Cn-0.2n=30-120C$, so $n(C-0.2)=30-120C$. The inverse is $n(C)=\frac{30-120C}{C-0.2}$.

Explanation

To find the inverse of a rational function, your goal is to free the variable from the fraction. After swapping $x$ and $y$, use algebra to get all $y$ terms on one side, factor out the $y$, and then divide to solve for it.