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5. Rational Exponents & Radical Functions
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Algebra 2

5.06 Inverse relations and functions

Lesson
Practice
Lesson

Concept summary

Inverse operations are operations that 'undo' each other - for example, addition and subtraction, or multiplication and division. We can extend this concept to find the inverse of an entire function. We can use function composition to check if a function is indeed an inverse of another.

Inverse function

Two functions, y=h\left(x\right) and x=g\left(y\right), are said to be inverses when g\left(h\left(x\right)\right)=x and h\left(g\left(y\right)\right)=y.

The function inverse to f\left(x\right) is denoted f^{-1}\left(x\right).

Inverse relation

A relation that reverses the original relation. The graph of an inverse relation is the original graph reflected across the line y=x.

We can find the inverse function algebraically by completing the following steps:

  1. Write f\left(x\right) as y
  2. Swap x with y and y with x
  3. Solve for y
  4. Replace y with f^{-1}\left(x\right) notation
  5. Check that the inverse is in fact a function using the vertical line test

Geometrically, swapping the x and y variables around means that the function and the inverse function are mirror images of each other across the line: y=x

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Swapping the x and y in a relationship will exchange the coordinates for any point. Thus, the domain and range will be switched in an inverse relation compared to the original relation.

It is important to note that all inverse functions are inverse relations, much like all functions are relations, but not all inverse relations are inverse functions. The inverse of a function is also a function only if, for every input, there is only one output. Such functions are called one-to-one functions.

One-to-one function

Functions that return a unique element in the range for each element in their domain.

A function such as f(x)=x^2 does not have an inverse function. If we reflect f(x)=x^2 across the line y=x we will get a relation that is not a function.

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We can see that the inverse relation x=y^2 or \\y=\pm \sqrt{x} does not pass the vertical line test - a straight line drawn vertically through the graph has more than one point of intersection. This means we can use a horizontal line test to determine whether a function is invertible over its domain.

If a function does not pass the horizontal line test, we can restrict its domain so that it does. It will then have an inverse function with a range corresponding to the restricted domain.

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If a function is not invertible over a particular domain, we can restrict the domain in such a way that the graph of the function passes the horizontal line test, and the function becomes invertible.

In the example above, if we restrict the domain of y=x^2 to x\geq0, it will now have an inverse function, namely y=\sqrt{x}.

Worked examples

Example 1

Consider the graphs of f\left(x\right), g\left(x\right) and h\left(x\right) and determine if they have an inverse function without any domain restrictions.

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a

f\left(x\right)

Approach

f\left(x\right) appears to be a linear function with a negative slope.

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  • We can graph the inverse relation by reflecting f\left(x\right) across the line y=x.
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  • We can then determine if the inverse relation is a function by drawing a vertical line through the function, and determining the number of times they intersect.
  • We can see that there is only one point of intersection.

All of this confirms that the original function f\left(x\right) was one-to-one.

Solution

We can see that f\left(x\right) has an inverse function since it is a one-to-one function.

Reflection

As the two functions are reflections of each other across the line y=x, we could instead apply the horizontal line test on f\left(x\right).

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  • If it only intersects in one place then there will be an inverse function, without any domain restrictions.
  • If it has more than one point of intersection there is not an inverse function
  • We can see that there is only one point of intersection.

The horizontal line test also tells us that f\left(x\right) is a one-to-one function. Since f\left(x\right) is a function and the horizontal line test tell us each point in the output is unique. Asking if a function "has an inverse function without any domain restrictions" is equivalent to asking if a function is one-to-one.

b

g\left(x\right)

Approach

g\left(x\right) appears to be a quadratic function opening upwards.

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  • We can graph the inverse relation by reflecting f\left(x\right) across the line y=x.
  • We can then determine if the inverse relation is a function by drawing a vertical line through the function, and determining the number of times they intersect.
  • We can see that there are two points of intersection.

All of this confirms that the original function g\left(x\right) was not a one-to-one function.

Solution

We can see that g\left(x\right) does not have an inverse function.

Reflection

Alternatively, we could perform the horizontal line test on g\left(x\right), revealing the function is not a one-to-one function.

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However, if g\left(x\right) had a restricted domain of either \left[0, \infty\right) or \left( -\infty, 0\right], then it would have an inverse function, as we can see below for \left[0, \infty\right).

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c

h\left(x\right)

Approach

h\left(x\right) appears to be a non-linear function with a negative slope.

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  • We can graph the inverse relation by reflecting h\left(x\right) across the line y=x.
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  • We can then determine if the inverse relation is a function by drawing a vertical line through the function, and determining the number of times they intersect.
  • We can see that there is only one point of intersection.

All of this confirms that the original function h\left(x\right) was one-to-one.

Solution

We can see that h\left(x\right) has an inverse function because it is a one-to-one function.

Reflection

As the two functions are reflections of each other across the line y=x, we could instead apply the horizontal line test on h\left(x\right).

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  • If it only intersects in one place then there will be an inverse function, without any domain restrictions.
  • If it has more than one point of intersection there is not an inverse function
  • We can see that there is only one point of intersection.

The horizontal line test also tells us that h\left(x\right) is a one-to-one function. Since h\left(x\right) is a function and the horizontal line test tell us each point in the output is unique.

Example 2

For each of the following functions:

  1. Determine an expression for the inverse relation.
  2. State whether or not the inverse is a function.
a

y = \dfrac{x }{3}+2

Approach

We want to swap x and y, and the solve for y. We can then check that the inverse is a function using the vertical line test.

Solution

\displaystyle x\displaystyle =\displaystyle \frac{y}{3}+2 Swap x and y
\displaystyle x-2\displaystyle =\displaystyle \frac{y}{3}Subtract 2 from both sides
\displaystyle 3\left(x-2\right)\displaystyle =\displaystyle yMultiply both sides by 3
\displaystyle 3x-6\displaystyle =\displaystyle yDistribute the parentheses
\displaystyle y\displaystyle =\displaystyle 3x-6Symmetric property of equality

y=3x-6 is the inverse of y=\dfrac{x}{3}+2.

We now need to determine if y=3x-6 is a function. Drawing a graph of y=3x-6, it is clear to see that it passes the vertical line test and therefore is a function.

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Reflection

Any straight line that is not vertical \left(x=c\right), is a function. This also means any straight line that is not horizontal \left(y=c\right), has an inverse function.

b

y = \left(x-3\right)^2-5

Approach

We want to swap x and y, and the solve for y. We can then check that the inverse is a function using the vertical line test.

Solution

\displaystyle x\displaystyle =\displaystyle \left(y-3\right)^2-5 Swap x and y
\displaystyle x+5\displaystyle =\displaystyle \left(y-3\right)^2Add 5 to both sides
\displaystyle \pm\sqrt{x+5}\displaystyle =\displaystyle y-3Take the square root of both sides
\displaystyle 3\pm\sqrt{x+5}\displaystyle =\displaystyle yAdd 3 to both sides

y=3\pm\sqrt{x+5} is the inverse of y=\left(x-3\right)^2-5.

We now need to determine if y=3\pm\sqrt{x+5} is a function. Drawing a graph of y=3\pm\sqrt{x+5}, it is clear to see that it does not pass the vertical line test and therefore is not a function.

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Reflection

As the graph of a quadratic function does not pass the horizontal line test, no quadratic functions have inverses without domain restrictions.

Example 3

For each of the following pairs of functions f and g, state whether or not the two functions are inverses:

a

f \left( x \right) = 5\left(x +4\right)^{3} and g \left( x \right) = \sqrt[3]{\dfrac{x}{5}}-4

Approach

We want to first substitute x=g\left(x\right) into f\left(x\right), and simplify the result. If f\left(g\left(x\right)\right)=x, we then do the opposite and substitute x=f\left(x\right) into g\left(x\right).

If the result of both these compositions is equal to x, then the two functions are inverses.

Solution

First let's find an expression for f\left(g\left(x\right)\right).

\displaystyle f\left(g\left(x\right)\right)\displaystyle =\displaystyle f\left( \sqrt[3]{\dfrac{x}{5}}-4\right)Substitute g \left( x \right) = \sqrt[3]{\dfrac{x}{5}}-4
\displaystyle =\displaystyle 5\left(\left( \sqrt[3]{\dfrac{x}{5}}-4\right) +4\right)^{3}Substitute f \left( x \right) = 5\left(x +4\right)^{3}
\displaystyle =\displaystyle 5\left( \sqrt[3]{\dfrac{x}{5}} \right)^{3}Combine like terms
\displaystyle =\displaystyle 5\left( \frac{x}{5}\right)\left(\sqrt[n]{a}\right)^n=a
\displaystyle =\displaystyle xSimplify

Now let's find an expression for g\left(f\left(x\right)\right).

\displaystyle g\left(f\left(x\right)\right)\displaystyle =\displaystyle g\left( 5\left(x +4\right)^{3}\right)Substitute f \left( x \right) = 5\left(x +4\right)^{3}
\displaystyle =\displaystyle \sqrt[3]{\frac{ 5\left(x +4\right)^{3}}{5}}-4Substitute g \left( x \right) = \sqrt[3]{\dfrac{x}{5}}-4
\displaystyle =\displaystyle \sqrt[3]{ \left(x +4\right)^{3}}-4Simplify the radicand
\displaystyle =\displaystyle \left(x +4\right)-4\left(\sqrt[n]{a}\right)^n=a
\displaystyle =\displaystyle xCombine like terms

As g\left(f\left(x\right)\right)=x=g\left(f\left(x\right)\right), the functions are inverses of each other.

Reflection

We must test that both g\left(f\left(x\right)\right) and f\left(g\left(x\right)\right) are equal to x to be sure that the functions are inverses of each other. However, we only need to disprove one of them to be sure that the functions are not inverses of each other.

b

f \left( x \right) = \dfrac{4}{x+5} and g \left( x \right) = \dfrac{-5x-4}{x}

Approach

We want to first substitute x=g\left(x\right) into f\left(x\right), and simplify the result. If f\left(g\left(x\right)\right)=x, we then do the opposite and substitute x=f\left(x\right) into g\left(x\right).

If the result of both these compositions is equal to x, then the two functions are inverses.

Solution

First let's find an expression for f\left(g\left(x\right)\right)

\displaystyle f\left(g\left(x\right)\right)\displaystyle =\displaystyle f\left( \dfrac{-5x-4}{x}\right)Substitute g \left( x \right) = \dfrac{-5x-4}{x}
\displaystyle =\displaystyle \frac{4}{\left( \dfrac{-5x-4}{x}\right)+5}Substitute f \left( x \right) = \dfrac{4}{x+5}
\displaystyle =\displaystyle \frac{4}{\left( \dfrac{-5x-4}{x}\right)+\dfrac{5x}{x}}Express 5 with a denominator of x
\displaystyle =\displaystyle \frac{4}{ \left(\dfrac{-4}{x}\right)}Simplify the denominator
\displaystyle =\displaystyle 4 \cdot \left(\frac{x}{-4}\right)Multiply by the reciprocal
\displaystyle =\displaystyle -xSimplify

As f\left(g\left(x\right)\right)\neq x, we do not need to check if g\left(f\left(x\right)\right)\neq x, as this is enough to prove that the functions are not inverses of each other.

Outcomes

A2.F.BF.B.4

Find the inverse of a function.

A2.F.BF.B.4.A

Determine whether a function is one-to-one.

A2.F.BF.B.4.B

Find the inverse of a function on an appropriate domain.

A2.F.BF.B.4.C

Given an invertible function on an appropriate domain, identify the domain of the inverse function.

A2.MP1

Make sense of problems and persevere in solving them.

A2.MP3

Construct viable arguments and critique the reasoning of others.

A2.MP6

Attend to precision.

A2.MP7

Look for and make use of structure.

A2.MP8

Look for and express regularity in repeated reasoning.

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