Problem 15 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems \(11-16\), find the inverse of the given function \(f\) and verify that \(f\left(f^{-1}(x)\right)=x\) for all \(x\) in the domain of \(f^{-1}\), and \(f^{-1}(f(x))=x\) for all \(x\) in the domain of \(f\). $$ f(x)=\log _{10}(3 x+2) $$

Step-by-Step Solution

Answer

The inverse function is \( f^{-1}(x) = \frac{10^x - 2}{3} \). This inverse satisfies both verification conditions.

Step 1: Express the Function in Terms of y

Start by expressing the function as an equation in terms of y: \[ y = \log_{10}(3x + 2) \].

Step 2: Convert to Exponential Form

Convert the logarithmic equation to exponential form to solve for x:\[ 10^y = 3x + 2 \].

Step 3: Solve for x

Rearrange the equation to isolate x:\[ 3x = 10^y - 2 \]\[ x = \frac{10^y - 2}{3} \].

Step 4: Express Inverse as a Function

Express the inverse function in terms of x by replacing y with x:\[ f^{-1}(x) = \frac{10^x - 2}{3} \].

Step 5: Verify the Composite Function f(f^{-1}(x))

Substitute \( f^{-1}(x) \) into \( f \):\[ f(f^{-1}(x)) = \log_{10} \left(3 \left(\frac{10^x - 2}{3}\right) + 2 \right) \]Simplifying gives:\[ = \log_{10}(10^x) = x \].

Step 6: Verify the Composite Function f^{-1}(f(x))

Substitute \( f(x) \) into \( f^{-1} \):\[ f^{-1}(f(x)) = \frac{10^{\log_{10}(3x + 2)} - 2}{3} \]Simplifying gives:\[ = \frac{3x + 2 - 2}{3} = x \].

Key Concepts

Logarithmic Functions

Logarithmic functions are the inverse of exponential functions. They allow us to solve equations where the unknown variable is an exponent. Consider a function like

• \( f(x) = \log_{10}(3x + 2) \)

For logarithmic functions, the base (in this case, 10) is important, as it determines how quickly the logarithm grows. This particular function shows how the expression \( 3x + 2 \) is transformed by the logarithm into a simpler form.
Logarithms are essential in many scientific disciplines for dealing with large ranges of values, such as the pH scale or decibel measurements. Understanding how to manipulate them is crucial for working with these sorts of scales.
For instance, converting the logarithmic function into an exponential function is a common tactic in solving equations. Moreover, a logarithmic function's domain is typically the range of its inverse, and vice versa. Understanding these relationships helps in determining the function's behavior and application.

Exponential Functions

Exponential functions are powerful mathematical tools defined by a constant base raised to a variable exponent. In the inverse function process, considering the function

• \( y = \log_{10}(3x + 2) \)

We express this logarithmic function as an exponential form:

• \( 10^y = 3x + 2 \)

This transformation is crucial because it allows us to solve for the variable \( x \). By expressing the log equation in terms of exponential form, we unlock the ability to rearrange and isolate terms.
Exponential functions expand rapidly, depicting scenarios such as population growth or radioactive decay. Understanding their properties aids in grasping the behavior of their inverse counterparts - logarithms. Moreover, knowing how to switch between these forms is fundamental in algebra and calculus, providing a pathway to solve real-world problems that involve growth and decay models.

Function Verification

Verifying inverse functions requires proving two main properties:

• That the composition of a function and its inverse returns the original input, \( f(f^{-1}(x)) = x \),
• and that the inverse composed with the function also yields the original input, \( f^{-1}(f(x)) = x \).

To verify these properties, consider:

• \( f(x) = \log_{10}(3x + 2) \)

and its inverse,

• \( f^{-1}(x) = \frac{10^x - 2}{3} \).

Substitute the inverse into the original function:

• \( f(f^{-1}(x)) = \log_{10}(10^x) = x \)

Then, perform the reverse by substituting the function into its inverse:

• \( f^{-1}(f(x)) = \frac{3x + 2 - 2}{3} = x \).

These steps prove the functions are truly inverses of each other. Function verification confirms that applying and then reversing these operations restores the original input. This validation process is fundamental in higher-level mathematics, ensuring mathematical functions' integrity and consistency.