Problem 33 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find a formula for \(f^{-1}(x)\) and then verify that \(f^{-1}(f(x))=x\) and \(f\left(f^{-1}(x)\right)=x\) $$ f(x)=\sqrt{x+1} $$

Step-by-Step Solution

Answer

The inverse is \( f^{-1}(x) = x^2 - 1 \). Both compositions verify the inverse.

Step 1: Define the Problem

The given function is \( f(x) = \sqrt{x + 1} \). Our task is to find its inverse, denoted \( f^{-1}(x) \), and then verify the inverse by showing that \( f^{-1}(f(x)) = x \) and \( f(f^{-1}(x)) = x \).

Step 2: Express the Function in Terms of y

Start by setting \( y = f(x) \), which gives us \( y = \sqrt{x + 1} \). Now express \( x \) in terms of \( y \).

Step 3: Solve for x

To find \( x \) in terms of \( y \), square both sides of the equation from the previous step: \( y^2 = x + 1 \). Then solve for \( x \) by subtracting 1 from both sides: \( x = y^2 - 1 \). This expression gives us the inverse function, \( f^{-1}(y) = y^2 - 1 \).

Step 4: Replace y with x for Inverse Function

To express the inverse function in terms of \( x \), replace \( y \) with \( x \). Thus, the inverse function is \( f^{-1}(x) = x^2 - 1 \).

Step 5: Verify Inverse with Composition

First verify that \( f^{-1}(f(x)) = x \). Substitute \( f(x) = \sqrt{x + 1} \) into \( f^{-1} \): \( f^{-1}(f(x)) = (\sqrt{x + 1})^2 - 1 = x + 1 - 1 = x \). This simplifies to confirm the identity.

Step 6: Verify Inverse with Reverse Composition

Now verify that \( f(f^{-1}(x)) = x \). Substitute \( f^{-1}(x) = x^2 - 1 \) into \( f \): \( f(f^{-1}(x)) = \sqrt{(x^2 - 1) + 1} = \sqrt{x^2} = x \). This simplifies to confirm the identity, assuming \( x \geq 0 \).

Key Concepts

Composition of Functions

The composition of functions is like preparing a multi-step recipe where the output of one function becomes the input for another. For a given function \( f(x) \) and its inverse function \( f^{-1}(x) \), the goal is to see how they interact. Here's what typically goes on in the composition of functions:

• Applying the Inverse: When you plug \( f(x) \) into \( f^{-1}(x) \), you get back to the same initial value \( x \). Mathematically, this is represented as \( f^{-1}(f(x)) = x \).
• Double Checking: If you reverse the process by putting \( f^{-1}(x) \) back into \( f(x) \), it should also simplify to \( x \). This is noted as \( f(f^{-1}(x)) = x \).

By confirming these identities, we establish that \( f\) and \( f^{-1}\) are indeed inverses, showcasing how compositions help verify functional relationships. This verification is crucial anytime you deal with inverse functions.

Verifying Inverse Functions

Verifying that a function has truly been inverted involves a bit of a dance with equations. After identifying the inverse, it's important to ensure that the original and inverse functions truly undo each other.The steps generally include:

• First Check: Take the composition \( f^{-1}(f(x)) \) and see if it simplifies to \( x \). With \( f(x) = \sqrt{x + 1} \) and \( f^{-1}(x) = x^2 - 1 \), substituting gives \( f^{-1}(\sqrt{x + 1}) = (\sqrt{x + 1})^2 - 1 = x \).
• Second Check: Do the reverse composition, \( f(f^{-1}(x)) \), and check that it also simplifies to \( x \). Using the inverses we found, this gives \( f(x^2 - 1) = \sqrt{(x^2 - 1) + 1} = \sqrt{x^2} = x \), assuming \( x \geq 0 \).

These two steps validate that the functions are indeed inverses, confirming the accuracy of the inversion process and ensuring no steps are missing.

Square Root Function

The square root function, represented as \( f(x) = \sqrt{x} \), is a specific type of radical function. It only returns non-negative results. This nature is important when considering inverses, particularly because an inverse should also reflect the domain and range adjustments.Key points to understand with square root functions:

• The Function's Domain: It is strictly the set of non-negative real numbers, \( x \geq 0 \). This is because you can't take the square root of a negative number without getting into imaginary numbers.
• Range: The output is also restricted to non-negative numbers. For \( f(x) = \sqrt{x+1} \), the domain starts from \( x = -1 \) (since \( x+1 \) needs to be non-negative), and the range is from \( 0 \) onward.
• Impact on Inversions: Since the output and input are both non-negative, any inverse functions particularly need to adhere to this rule, limiting \( x \) to values where the square root is defined.

These features make sure that when reversing a square root function, the values will reflect the appropriate real-world scenarios we're modeling.