# 6.1: Inverse Functions

**At Grade**Created by: CK-12

Functions such as logarithms, exponential functions, and trigonometric functions are examples of ** transcendental functions.** If a function is transcendental, it cannot be expressed as a polynomial or rational function. That is, it is not an

*algebraic function.*In this chapter, we will begin by developing the concept of an inverse of a function and how it is linked to its original numerically, algebraically, and graphically. Later, we will take each type of elementary transcendental function—logarithmic, exponential, and trigonometric—individually and see the connection between them and their respective inverses, derivatives, and integrals.

## Learning Objectives

A student will be able to:

- Understand the basic properties of the inverse of a function and how to find it.
- Understand how a function and its inverse are represented graphically.
- Know the conditions of invertabilty of a function.

### One-to-One Functions

A function, as you know from your previous mathematics background, is a rule that assigns a single value in its range to each point in its domain. In other words, for each output number, there is one or more input numbers. However, a function never produces more than a single output for one input. A function is said to be a ** one-to-one** function if each output is associated with only one single input. For example, \begin{align*}f(x) = x^2\end{align*} assigns the output \begin{align*}9\end{align*} for both \begin{align*}3\end{align*} and \begin{align*}-3,\end{align*} and thus it is not a

*one-to-one*function.

**One-to-One Function**

The function \begin{align*}f(x)\end{align*} is one-to-one in a domain \begin{align*}D\end{align*} if \begin{align*}f(a) \neq f(b),\end{align*} whenever \begin{align*}a \neq b.\end{align*}

There is an easy method to check if a function is one-to-one: draw a horizontal line across the graph. If the line intersects at only one point on the graph, then the function is one-to-one; otherwise, it is not. Notice in the figure below that the graph of \begin{align*}y = x^2\end{align*} is not one-to-one since the horizontal line intersects the graph more than once. But the function \begin{align*}y = x^3\end{align*} is a one-to-one function because the graph meets the horizontal line only once.

**Example 1:**

Determine whether the functions are one-to-one: (a) \begin{align*}f(x) = |x|\end{align*} (b) \begin{align*}h(x) = x^{1/2}.\end{align*}

*Solution:*

It is best to graph both functions and draw on each a horizontal line. As you can see from the graphs, \begin{align*}f(x) = |x|\end{align*} is not one-to-one since the horizontal line intersects it at two points. The function \begin{align*}h(x) = x^{1/2},\end{align*} however, is indeed one-to-one since only one point is intersected by the horizontal line.

### The Inverse of a Function

We discussed above the condition for a one-to-one function: for each output, there is only one input. A one-to-one function can be reversed in such a way that the input of the function becomes the output and the output becomes an input. This reverse of the original function is called the *inverse* of the function. If \begin{align*} f^{-1} \end{align*} is an inverse of a function \begin{align*}f,\end{align*} then \begin{align*} f^{-1} \circ f = x.\end{align*} For example, the two functions \begin{align*} f(x)=2x+3 \end{align*} and \begin{align*} h(x) = \frac{x-3}{2} \end{align*} are inverses of each other since

\begin{align*}f \circ h & = f(h(x)) = 2 \left [ \frac{x-3}{2} \right ] +3 = x-3+3 =x,\\ h \circ f & = h(f(x)) = \frac {(2x+3)-3}{2} = \frac {2x}{x} = x.\end{align*}

Thus

\begin{align*}f \circ h = h \circ f = x,\end{align*}

and \begin{align*}f\end{align*} and \begin{align*}h\end{align*} are inverses of each other.

Note: In general, \begin{align*} f^{-1} \neq \frac{1}{f}.\end{align*}

**When is a function invertable?**

It is interesting to note that if a function \begin{align*}f(x)\end{align*} is always increasing or always decreasing over its domain, then a horizontal line will cut through this graph at one point only. Then \begin{align*}f\end{align*} in this case is a one-to-one function and thus has an inverse. So if we can find a way to prove that a function is constantly increasing or decreasing, then it is ** invertable** or

**. From previous chapters, you have learned that if \begin{align*}f'(x) > 0\end{align*} then \begin{align*}f\end{align*} must be increasing and if \begin{align*}f'(x) < 0\end{align*} then \begin{align*}f\end{align*} must be decreasing.**

*monotonic*To summarize, a function has an inverse if it is one-to-one in its domain or if its derivative is either \begin{align*}f'(x) > 0\end{align*} or \begin{align*}f'(x) < 0.\end{align*}

**Example 2:**

Given the polynomial function \begin{align*}f(x) = 3x^5 + 2x + 1,\end{align*} show that it is invertable (has an inverse).

*Solution:*

Taking the derivative, we find that \begin{align*}f'(x) = 15x^4 + 2 > 0\end{align*} for all \begin{align*}x.\end{align*} We conclude that \begin{align*}f(x)\end{align*} is one-to-one and invertable. Keep in mind that it may not be easy to find the inverse of \begin{align*}f(x) = 3x^5 + 2x + 1\end{align*} (try it!), but we still know that it is indeed invertable.

**How to find the inverse of a one-to-one function:**

To find the inverse of a one-to-one function, simply solve for \begin{align*}x\end{align*} in terms of \begin{align*}y\end{align*} and then interchange \begin{align*}x\end{align*} and \begin{align*}y.\end{align*} The resulting formula is the inverse \begin{align*}y = f^{-1}(x).\end{align*}

**Example 3:**

Find the inverse of \begin{align*} f(x) = \sqrt {4x + 1}\end{align*}.

*Solution:*

From the discussion above, we can find the inverse by first solving for \begin{align*} x\end{align*} in \begin{align*} y = \sqrt {4x + 1}\end{align*}.

\begin{align*}y & = \sqrt {4x + 1},\\ y^2 & = 4x + 1,\\ x & = \frac { y^2 - 1}{4}.\end{align*}

Interchanging \begin{align*} x \longleftrightarrow y\end{align*},

\begin{align*} y = \frac {x^2 - 1}{4}.\end{align*}

Replacing \begin{align*} y = f^{-1}(x),\end{align*}

\begin{align*} f^{-1}(x) = \frac {x^2 - 1}{4}\end{align*}

which is the inverse of the original function \begin{align*} f(x) = \sqrt {4x + 1} \end{align*}.

### Graphs of Inverse Functions

What is the relationship between the graphs of \begin{align*}f\end{align*} and \begin{align*}f^{-1}\end{align*}? If the point \begin{align*}(a,b)\end{align*} is on the graph of \begin{align*}f(x),\end{align*} then from the definition of the inverse, the point \begin{align*}(b, a)\end{align*} is on the graph of \begin{align*}f^{-1}(x).\end{align*} In other words, when we reverse the coordinates of a point on the graph of \begin{align*}f(x)\end{align*} we automatically get a point on the graph of \begin{align*}f^{-1}(x).\end{align*} We conclude that \begin{align*}f(x)\end{align*} and \begin{align*}f^{-1}(x)\end{align*} are ** reflections** of one another about the line \begin{align*}y = x.\end{align*} That is, each is a mirror image of the other about the line \begin{align*}y = x.\end{align*} The figure below shows an example of \begin{align*}y=x^2\end{align*} and, when the domain is restricted, its inverse \begin{align*}y = \sqrt{x}\end{align*} and how they are reflected about \begin{align*}y = x\end{align*}.

It is important to note that for the function \begin{align*}f(x) = x^2\end{align*} to have an inverse, we must restrict its domain to \begin{align*} 0 \le x < \infty,\end{align*} since that is the domain in which the function is increasing.

## Continuity and Differentiability of Inverse Functions

Since the graph of a one-to-one function and its inverse are reflections of one another about the line \begin{align*}y = x,\end{align*} it would be safe to say that if the function \begin{align*}f\end{align*} has no breaks (no discontinuities) then \begin{align*}f^{-1}\end{align*} will not have breaks either. This implies that if \begin{align*}f\end{align*} is continuous on the domain \begin{align*}D,\end{align*} then its inverse \begin{align*}f^{-1}\end{align*} is continuous on the range \begin{align*}R\end{align*} of \begin{align*}f.\end{align*} For example, if \begin{align*} f(x) = \sqrt{x}\end{align*}, then its domain is \begin{align*}x \ge 0\end{align*} and its range is \begin{align*}y \ge 0.\end{align*} This means that \begin{align*}f(x)\end{align*} is continuous for all \begin{align*}x \ge 0.\end{align*} The inverse of \begin{align*}f(x)\end{align*} is \begin{align*}f^{-1}(x) = x^2,\end{align*} where its domain is all \begin{align*}x > 0\end{align*} and its range is \begin{align*}y \ge 0.\end{align*} We conclude that if \begin{align*}f\end{align*} is a function with domain \begin{align*}D\end{align*} and range \begin{align*}R\end{align*} and it is continuous and one-to-one on \begin{align*}D,\end{align*} then its inverse \begin{align*}f^{-1}\end{align*} is continuous and one-to-one on the range \begin{align*}R\end{align*} of \begin{align*}f.\end{align*}

Suppose that \begin{align*}f\end{align*} has a domain \begin{align*}D\end{align*} and a range \begin{align*}R.\end{align*} If \begin{align*}f\end{align*} is differentiable and one-to-one on \begin{align*}D,\end{align*} then its inverse \begin{align*}f^{-1}\end{align*} is differentiable at any value \begin{align*}x\end{align*} in \begin{align*}R\end{align*} for which \begin{align*}f'(f^{-1}(x)) \neq 0\end{align*} and

\begin{align*} \frac {d}{dx} [f^{-1}(x)] = \frac{1} {f'(f^{-1}(x))}.\end{align*}

The formula above can be written in a form that is easier to remember:

\begin{align*} \frac{dy} {dx} = \frac{1} {dx / dy}.\end{align*}

In addition, if \begin{align*}f\end{align*} on its domain is either \begin{align*}f'(x) > 0\end{align*} or \begin{align*}f'(x) < 0,\end{align*} then \begin{align*}f\end{align*} has an inverse function \begin{align*}f^{-1}\end{align*} and \begin{align*}f^{-1}\end{align*} is differentiable at all values of \begin{align*}x\end{align*} in the range of \begin{align*}f.\end{align*} In this case, \begin{align*}f^{-1}\end{align*} is given by the formula above. The example below illustrate this important theorem.

**Example 4:**

In Example 3, we were given the polynomial function \begin{align*}f(x) = 3x^5 + 2x + 1\end{align*} and we showed that it is invertable. Show that it is differentiable and find the derivative of its inverse.

**Solution:**

Since \begin{align*}f'(x) = 15x^4 + 2 > 0\end{align*} for all \begin{align*}x \in R,\end{align*} \begin{align*}f^{-1}(x)\end{align*} is differentiable at all values of \begin{align*}x.\end{align*} To find the derivative of \begin{align*}f^{-1},\end{align*} if we let \begin{align*}x = f(y),\end{align*} then

\begin{align*} x = f(y) = 3y^5 + 2y + 1.\end{align*}

So

\begin{align*} \frac{dx} {dy} = 15y^4 + 2\end{align*}

and

\begin{align*} \frac{dy} {dx} = \frac{1} {dx/dy} = \frac{1} {15y^4 + 2}.\end{align*}

Since we are unable to solve for \begin{align*}y\end{align*} in terms of \begin{align*}x,\end{align*} we leave the answer above in terms of \begin{align*}y.\end{align*} Another way of solving the problem is to use Implicit Differentiation:

Since

\begin{align*}x = 3y^5+ 2y + 1,\end{align*}

differentiating implicitly,

\begin{align*}\frac{d}{dx} [x] & = \frac{d}{dx} [3y^5 + 2y + 1],\\ 1 & = (15y^4 + 2) \frac{dy} {dx}.\end{align*}

Solving for \begin{align*} \frac{dy}{dx}\end{align*} we finally obtain

\begin{align*} \frac{dy}{dx} = \frac{1} {15y^4 + 2},\end{align*}

which is the same result.

## Review Questions

In problems #1 - 3, find the inverse function of \begin{align*}f\end{align*} and verify that \begin{align*}f \circ f^{-1} = f^{-1} \circ f = x.\end{align*}

- \begin{align*}f(x) = 3x + 1\end{align*}
- \begin{align*} \sqrt[3]{x}\end{align*}
- \begin{align*} f(x) = \frac{x - 1} {3}\end{align*}

In problems #4 - 6, use the horizontal line test to verify whether the following functions have inverse.

- \begin{align*} h(x) = \frac{4 - x} {6}\end{align*}
- \begin{align*}g(x) = |x + 4| - |x - 4|\end{align*}
- \begin{align*}f(x) = -2x \sqrt{16 - x^2}\end{align*}

In problems #7 - 8, use the functions \begin{align*}f(x) = x + 4\end{align*} and \begin{align*}g(x) = 2x - 5\end{align*} to find the specified functions.

- \begin{align*}g^{-1} \circ f^{-1}\end{align*}
- \begin{align*}(f \circ g)^{-1}\end{align*}

In problems #9 - 10, show that \begin{align*}f\end{align*} is monotonic (invertable) on the given interval (and therefore has an inverse.)

- \begin{align*}f(x) = (x - 5)^2, [5, \infty)\end{align*}
- \begin{align*}f(x) = \cos x, \left [0, \frac{\pi} {2}\right ]\end{align*}

## Review Answers

- \begin{align*} f^{-1}(x) = \frac{x - 1} {3}\end{align*}
- \begin{align*}x^3\end{align*}
- \begin{align*}3x + 1\end{align*}
- Function has an inverse.
- Function does not have an inverse.
- Function does not have an inverse.
- \begin{align*} \frac{x + 1} {2}\end{align*}
- \begin{align*} \frac{x + 1} {2}\end{align*}
- \begin{align*}f'(x) = 2(x - 5) > 0\end{align*} on \begin{align*}x > 5.\end{align*}
- \begin{align*} f'(x) = -\sin x,\end{align*} which is negative on the interval in question, so \begin{align*}f(x)\end{align*} is monotonically decreasing.

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