AP Precalculus – Inverse Functions and Everything You Need to Know

Inverse Functions

A function \(f(x) = x + 3 \) from \(A = \{1,\, 2,\, 3,\, 4\} \) to \(B = \{4,\, 5,\, 6,\, 7\} \) can be written as:

\(f : \{(1,\, 4),\, (2,\, 5),\, (3,\, 6),\, (4,\, 7)\} \).

The inverse function \(f^{-1} \) is formed by interchanging the coordinates: \(f^{-1} : \{(4,\, 1),\, (5,\, 2),\, (6,\, 3),\, (7,\, 4)\} \).

The domain of \(f \) equals the range of \(f^{-1} \) and vice versa.

\(f(f^{-1}(x)) = x \) and \(f^{-1}(f(x)) = x \).

Inverse Function Definition

\(f(g(x)) = x \) and \(g(f(x)) = x \).

The inverse of \(f \) is denoted \(f^{-1} \).

Key Observations

1. If \(g \) is the inverse of \(f \), then \(f \) is the inverse of \(g \).
2. The domain of \(f^{-1} \) equals the range of \(f \), and the range of \(f^{-1} \) equals the domain of \(f \).
3. Not all functions have inverses, but when they do, it’s unique.

Subtraction undoes addition: \(f(x) = x + c \), \(f^{-1}(x) = x – c \).

Division undoes multiplication: \(f(x) = cx \), \(f^{-1}(x) = \frac{x}{c} \), \(c \neq 0 \).

Example 1: Verifying Inverses

Prove \(f(x) = 2x^3 – 1 \) and \(g(x) = \sqrt[3]{\frac{x + 1}{2}} \) are inverses:

\(f(g(x)) = x \) and \(g(f(x)) = x \), confirming they are inverses.

Theorem 1: Reflective Property

If \(f \) has the point \((a, b) \), then \(f^{-1} \) contains \((b, a) \).

Theorem 2: Existence of an Inverse

1. A function has an inverse if and only if it is one-to-one.
2. If \(f \) is strictly monotonic, it is one-to-one and has an inverse.

Example 2: Inverse Function Existence

For \(f(x) = x^3 + x – 1 \), since \(f'(x) = 3x^2 + 1 > 0 \), \(f \) is strictly monotonic and has an inverse.

For \(f(x) = x^3 – x + 1 \), it fails the Horizontal Line Test, so no inverse exists.

Steps to Find an Inverse Function

1. Check if \(f \) is one-to-one using Theorem 2.
2. Solve for \(x \) in terms of \(y \): \(x = f^{-1}(y) \).
3. Interchange \(x \) and \(y \) to find \(y = f^{-1}(x) \).
4. Define the domain of \(f^{-1} \).
5. Verify \(f(f^{-1}(x)) = x \) and \(f^{-1}(f(x)) = x \).

Example 3: Finding Inverses

For \(f(x) = \sqrt{2x – 3} \), solve for \(x \) in terms of \(y \) to get \(f^{-1}(x) = \frac{x^2 + 3}{2} \).

Solution: From the graph of \(f \), it appears that \(f \) is increasing over its entire domain, \([3/2, \infty) \).

So, \(f \) is strictly monotonic, and it must have an inverse function.

To find an equation for the inverse function, let \(y = f(x) \), and solve for \(x \) in terms of \(y \):

\(\sqrt{2x – 3} = y \)     Let y = f(x)

\(2x – 3 = y^2 \)     Square each side

\(x = \frac{y^2 + 3}{2} \)     Solve for x

\(y = \frac{x^2 + 3}{2} \)     Interchange x and y

\(f^{-1}(x) = \frac{x^2 + 3}{2} \)     Replace y by f^-1(x)

The domain of \(f^{-1} \) is the range of \(f \), which is \([0, \infty) \).

You can verify this result as shown below:

\(f(f^{-1}(x)) = \sqrt{2 \left( \frac{x^2 + 3}{2} \right) – 3} = \sqrt{x^2} = x, \quad x \geq 0 \)

\(f^{-1}(f(x)) = \left( \frac{\sqrt{2x – 3}}{2} \right)^2 + 3 = \frac{2x – 3 + 3}{2} = x, \quad x \geq \frac{3}{2} \)

Example 4: Testing One-to-One

For \(f(x) = \sin x \), it’s not one-to-one on \((-\infty, \infty) \), but is one-to-one on \([-\pi/2, \pi/2] \).

Example 5

If \(f(1) = 5 \), \(f(3) = 7 \), and \(f(8) = -10 \), find \(f^{-1}(5) \), \(f^{-1}(7) \), and \(f^{-1}(-10) \).

Solution:

From the definition of \(f^{-1} \), we have:

\(f^{-1}(5) = 1 \) because \(f(1) = 5 \)

\(f^{-1}(7) = 3 \) because \(f(3) = 7 \)

\(f^{-1}(-10) = 8 \) because \(f(8) = -10 \)

This shows how \(f^{-1} \) reverses the effect of \(f \) in this case.