Problem 11 - 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 \(f\). $$ f(x)=\frac{1}{1+2^{x}} $$

Step-by-Step Solution

Answer

The inverse is \(f^{-1}(x) = \log_2 \left( \frac{1-x}{x} \right)\); both inverse properties are verified.

Step 1: Express the function in terms of y

To find the inverse, first express the function in terms of \(y\). This gives us \(y = \frac{1}{1 + 2^x}\).

Step 2: Solve for x

Next, solve the equation from Step 1 for \(x\). Start by multiplying both sides by \(1 + 2^x\):\[ y(1 + 2^x) = 1 \]Rearrange to:\[ y + y \, 2^x = 1 \]Now solve for \(2^x\):\[ 2^x = \frac{1 - y}{y} \]

Step 3: Express x in terms of y using logarithms

Take the logarithm of both sides to isolate \(x\):\[ x = \log_2 \left( \frac{1-y}{y} \right) \]

Step 4: Swap x and y for the inverse function

Now, exchange \(x\) and \(y\) to write the inverse function:\[ f^{-1}(x) = \log_2 \left( \frac{1-x}{x} \right) \]

Step 5: Verify inverse property 1

To verify \(f(f^{-1}(x)) = x\), substitute \(f^{-1}(x)\) into \(f\):\[ f(f^{-1}(x)) = \frac{1}{1 + 2^{\log_2 \left( \frac{1-x}{x} \right)}} \]Simplify:\[ 2^{\log_2 \left( \frac{1-x}{x} \right)} = \frac{1-x}{x} \]Hence:\[ f(f^{-1}(x)) = \frac{1}{1 + \frac{1-x}{x}} = \frac{x}{1} = x \]

Step 6: Verify inverse property 2

To verify \(f^{-1}(f(x)) = x\), substitute \(f(x)\) into \(f^{-1}\):\[ f^{-1}(f(x)) = \log_2 \left( \frac{1 - \frac{1}{1 + 2^x}}{\frac{1}{1 + 2^x}} \right) \]Simplify:\[ \frac{1 - \frac{1}{1 + 2^x}}{\frac{1}{1 + 2^x}} = 2^x \]Thus:\[ f^{-1}(f(x)) = \log_2(2^x) = x \]

Step 7: Conclusion

Since both properties \(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x\) are verified, the inverse function is correctly found. The inverse function is \(f^{-1}(x) = \log_2 \left( \frac{1-x}{x} \right)\).

Key Concepts

Function Verification

Function verification is key to understanding how functions and their inverses relate to one another. When you find the inverse of a function, you're essentially "undoing" what the original function does. For this process, we verify that the inverse function is correct by using two important properties:

• The first property is that when you apply the original function and then the inverse function to any value in the domain, you should get back the original value: \( f(f^{-1}(x)) = x \).
• The second property is similar but in reverse order: \( f^{-1}(f(x)) = x \). This means that applying the inverse and then the original function returns the value you started with.

By verifying both of these conditions, we can confirm that the inverse function is correctly derived and accurate.

Logarithms

Logarithms are the inverse operation of exponentiation. They're extremely useful when solving equations that involve exponential terms, as they allow us to convert multiplication into addition, making the equation easier to handle.In our problem, to isolate \(x\) from the term \(2^x\), we use logarithms. Specifically, we take the base 2 logarithm (\(\log_2\)) because it directly relates to the base of the exponent, 2. The use of the logarithm transforms the exponential equation into a linear one, simplifying the process and allowing us to solve for the variable.Using logarithms effectively often involves understanding their properties and rules, such as:

• The product rule: \(\log_b(MN) = \log_b(M) + \log_b(N) \)
• The change of base formula, which allows you to compute a logarithm in terms of logs written with another base
• Understanding the relationship between logs and exponents: for example, \(b^{\log_b(x)} = x \).

Solving Equations

Solving equations, especially those involving functions and their inverses, requires methodical steps. In this exercise, solving for \(x\) involves manipulating an equation through various algebraic techniques. We started by writing the original equation in terms of \(y\), set equal to \(f(x)\). Then, we rearranged it to solve for \(x\) by first getting rid of any fractions or unnecessary terms. This involved multiplying both sides and then rearranging to isolate terms involving \(x\).Once the equation was rearranged, we used logarithms to solve for \(x\), which were crucial in handling exponential terms, as discussed. Finally, to find the inverse, swapping \(x\) and \(y\) was necessary, finalizing our expression for the inverse function.These steps reiterated key skills in equation solving:

• Rearranging terms to isolate the unknown variable
• Using mathematical operations like multiplication or division to simplify
• Applying inverse operations, such as using a logarithm to undo an exponentiation.

Mastering these sorts of problems involves practice and a good understanding of both algebraic manipulation and the properties of logarithms.