Question

In general, two functions \(f(x)\) and \(g(x)\) are inverses of each other if and only if \(f(g(x))=x\) and \(g(f(x))=x .\) Verify that the pairs of functions are inverses of each other. a) \(f(x)=5 x+10\) and \(g(x)=\frac{1}{5} x-2\) b) \(f(x)=\frac{x-1}{2}\) and \(g(x)=2 x+1\) c) \(f(x)=\sqrt[3]{x+1}\) and \(g(x)=x^{3}-1\) d) \(f(x)=5^{x}\) and \(g(x)=\log _{5} x\)

Short answer

All given pairs of functions are inverses because they satisfy \(f(g(x)) = x\) and \(g(f(x)) = x\).

Step-by-step solution

- Verify for the pair (a)

First, substitute the function g(x) into f(x):\[f(g(x)) = f\left( \frac{1}{5}x - 2 \right)\]\[= 5\left(\frac{1}{5}x - 2\right) + 10 \]\[= x - 10 + 10 \]\[= x\]Now, substitute the function f(x) into g(x):\[g(f(x)) = g(5x + 10)\]\[= \frac{1}{5}(5x + 10) - 2\]\[= x + 2 - 2\]\[= x\]Since both conditions are satisfied, \(f(x) = 5x + 10\) and \(g(x) = \frac{1}{5}x - 2\) are inverses.

- Verify for the pair (b)

First, substitute the function g(x) into f(x):\[f(g(x)) = f(2x + 1)\]\[= \frac{2x + 1 - 1}{2} \]\[= \frac{2x}{2} \]\[= x\]Next, substitute the function f(x) into g(x):\[g(f(x)) = g\left( \frac{x - 1}{2} \right) \]\[= 2\left( \frac{x - 1}{2} \right) + 1 \]\[= x - 1 + 1 \]\[= x\]Since both conditions are satisfied, \(f(x) = \frac{x-1}{2}\) and \(g(x) = 2x + 1\) are inverses.

- Verify for the pair (c)

First, substitute the function g(x) into f(x):\[f(g(x)) = f(x^3 - 1)\]\[= \sqrt[3]{(x^3 - 1) + 1} \]\[= \sqrt[3]{x^3} \]\[= x\]Next, substitute the function f(x) into g(x):\[g(f(x)) = g(\sqrt[3]{x+1})\]\[= (\sqrt[3]{x+1})^3 - 1 \]\[= x + 1 - 1 \]\[= x\]Since both conditions are satisfied, \(f(x) = \sqrt[3]{x+1}\) and \(g(x) = x^3 - 1\) are inverses.

- Verify for the pair (d)

First, substitute the function g(x) into f(x):\[f(g(x)) = f(\log_5 x)\]\[= 5^{\log_5 x} \]\[= x\]Next, substitute the function f(x) into g(x):\[g(f(x)) = g(5^x)\]\[= \log_5(5^x)\]\[= x\]Since both conditions are satisfied, \(f(x) = 5^x\) and \(g(x) = \log_5 x\) are inverses.

Key concepts

Function Composition

To understand inverse functions, it's essential to know about function composition. Function composition involves applying one function to the results of another function. If you have two functions, say \(f(x)\) and \(g(x)\), you can create a new function by composing them in two ways: \(f(g(x))\) and \(g(f(x))\). Essentially, you substitute one function into the other.
For example, consider \(f(x) = 2x + 3\) and \(g(x) = x - 1\). To find \(f(g(x))\), substitute \(g(x)\) into \(f(x)\):

• Step 1: Substitute \(g(x)\) into \(f(x)\): \(f(g(x)) = f(x - 1)\)
• Step 2: Calculate the result: \(f(x - 1) = 2(x - 1) + 3 = 2x - 2 + 3 = 2x + 1\)

Similarly, for \(g(f(x))\):

• Step 1: Substitute \(f(x)\) into \(g(x)\): \(g(f(x)) = g(2x + 3)\)
• Step 2: Calculate the result: \(g(2x + 3) = (2x + 3) - 1 = 2x + 2\)

Function composition helps verify whether two functions are inverses.

Precalculus Verification

Precalculus verification is the process of confirming that two functions are indeed inverses. This involves checking two important conditions: \(f(g(x)) = x\) and \(g(f(x)) = x\). If both conditions hold true, the functions are inverses.
Let’s verify the pairs of functions provided in the exercise:
a) \(f(x) = 5x + 10\) and \(g(x) = \frac{1}{5}x - 2\)

• Substitute \(g(x)\) into \(f(x)\): \(f(g(x)) = f\left( \frac{1}{5}x - 2 \right) = 5\left(\frac{1}{5}x - 2 \right) + 10 = x\)
• Substitute \(f(x)\) into \(g(x)\): \(g(f(x)) = g(5x + 10) = \frac{1}{5}(5x + 10) - 2 = x\)

Since both conditions are satisfied, these functions are inverses.
We repeat similar verification steps for pairs b, c, and d as detailed in the original solution.

Inverse Properties

Inverse functions have unique properties that distinguish them. If \(f\) and \(g\) are inverses, applying one function after the other returns the starting value: \(f(g(x)) = x\) and \(g(f(x)) = x\). This cancellation effect is fundamental to the concept of inverse functions.
Consider a real-life analogy: If you lock a door (apply \(f\)), then unlock it (apply \(g\)), you return to the initial state of the door being unlocked.
This cancellation leads to several important implications:

• One-to-One Functions: Only one-to-one functions (bijective functions) have inverses. These functions pass the horizontal line test.
• Symmetry: If \(y = f(x)\), then \(x = g(y)\). The graph of \(f\) and its inverse \(g\) will be symmetric with respect to the line \(y = x\).
• Notational Clarity: The inverse of a function \(f\) is denoted \(f^{-1}\), but this notation does not imply exponentiation.

Understanding these properties provides deeper insights into how inverse functions operate and their significance in mathematics.