Problem 72 - Algebra for College Students

Question

Use this approach to find the inverse of each of the following functions. See below. (a) \(f(x)=3 x-9\) (b) \(f(x)=-2 x+6\) (c) \(f(x)=-x+1\) (d) \(f(x)=2 x\) (e) \(f(x)=-5 x\) (f) \(f(x)=x^{2}+6\) for \(x \geq 0\) If \(f(x)=2 x+3\) and \(g(x)=3 x-5\), find (a) \((f \circ g)^{-1}(x) \frac{x+7}{6}\) (b) \(\left(f^{-1} \circ g^{-1}\right)(x)\) (c) \(\left(g^{-1} \circ f^{-1}\right)(x)\) \(\frac{x-4}{6}\)

Step-by-Step Solution

Answer

The inverses are: (a) \(\frac{x+9}{3}\), (b) \(\frac{-x+6}{2}\), (c) \(1-x\), (d) \(\frac{x}{2}\), (e) \(-\frac{x}{5}\), (f) \(\sqrt{x-6}\). For compositions: (a) \(\frac{x+7}{6}\), (b) \(\frac{x-4}{6}\), (c) \(\frac{x-4}{6}\).

Step 1: Determine the Inverse for f(x)=3x-9

To find the inverse of the function, start by replacing \(f(x)\) with \(y\): \(y = 3x - 9\). Swap \(x\) and \(y\) to solve for the inverse: \(x = 3y - 9\). Solve for \(y\) by isolating it on one side of the equation: \(x + 9 = 3y\), and then \(y = \frac{x + 9}{3}\). Thus, the inverse function is \(f^{-1}(x) = \frac{x + 9}{3}\).

Step 2: Determine the Inverse for f(x)=-2x+6

Start by setting \(y = f(x) = -2x + 6\). Swap \(x\) and \(y\): \(x = -2y + 6\). Solve for \(y\): \(x - 6 = -2y\), and rearrange to get \(y = -\frac{x - 6}{2}\). Simplifying gives the inverse function \(f^{-1}(x) = -\frac{x - 6}{2} = \frac{-x + 6}{2}\).

Step 3: Determine the Inverse for f(x)=-x+1

Set \(y = -x + 1\) and swap \(x\) and \(y\): \(x = -y + 1\). Solve for \(y\): \(y = 1 - x\). The inverse function is then \(f^{-1}(x) = 1 - x\).

Step 4: Determine the Inverse for f(x)=2x

Let \(y = 2x\) and swap variables: \(x = 2y\). Solving for \(y\) gives \(y = \frac{x}{2}\). Therefore, the inverse function is \(f^{-1}(x) = \frac{x}{2}\).

Step 5: Determine the Inverse for f(x)=-5x

With \(y = -5x\) and after swapping variables to get \(x = -5y\), solve for \(y\): \(y = -\frac{x}{5}\). Hence, the inverse function is \(f^{-1}(x) = -\frac{x}{5}\).

Step 6: Determine the Inverse for f(x)=x^2+6, x>=0

For \(y = x^2 + 6\), substitute and swap: \(x = y^2 + 6\). We then have \(x - 6 = y^2\), so \(y = \sqrt{x - 6}\). The inverse function for \(x \geq 0\) is \(f^{-1}(x) = \sqrt{x - 6}\).

Step 7: Find (f ∘ g)^{-1}(x) given f(x)=2x+3 and g(x)=3x-5

Compute \((f \circ g)(x) = f(g(x))\): \((f \circ g)(x) = 2(3x - 5) + 3\). Simplify: \((f \circ g)(x) = 6x - 10 + 3 = 6x - 7\). Replace \(f \circ g(x)\) with \(y\): \(y = 6x - 7\). Swap \(x\) and \(y\) to solve for \(y\): \(x = 6y - 7\), thus, \(y = \frac{x + 7}{6}\). Hence, \( (f \circ g)^{-1}(x) = \frac{x + 7}{6}\).

Step 8: Find \\(\left(f^{-1} \\circ g^{-1}\right)(x)\\) with f(x)=2x+3 and g(x)=3x-5

First find \(f^{-1}(x)\): start with \(y = 2x + 3\), swap to get \(x = 2y + 3\), solve for \(y\): \(y = \frac{x - 3}{2}\). Now find \(g^{-1}(x)\): start with \(y = 3x - 5\), swap to get \(x = 3y - 5\), solve for \(y\): \(y = \frac{x + 5}{3}\). Therefore, the composition is \((f^{-1} \circ g^{-1})(x) = f^{-1}(g^{-1}(x)) = \frac{\left(\frac{x + 5}{3}\right) - 3}{2}\), simplifying gives \(\frac{x - 4}{6}\).

Step 9: Find \\(\left(g^{-1} \circ f^{-1}\right)(x)\\) given the same functions

Use previously found inverses: \(f^{-1}(x) = \frac{x - 3}{2}\) and \(g^{-1}(x) = \frac{x + 5}{3}\). Then \((g^{-1} \circ f^{-1})(x) = g^{-1}(f^{-1}(x)) = \frac{\left(\frac{x - 3}{2}\right) + 5}{3}\). Simplifying gives \(\frac{x - 4}{6}\).

Key Concepts

Function Composition

Function composition involves creating a new function by combining two existing functions. When you see notation like \(f \circ g\), it represents the function \(f(g(x))\). What this means is that for every value of \(x\), you first apply the function \(g(x)\) and then take the result and apply \(f\) to it. This can be an essential skill in understanding how different functions interact with one another.

• Start with the innermost function: Solve for \(g(x)\).
• Use the output of \(g(x)\) as the input for \(f(x)\).
• The end result is a composition, sometimes resulting in a more complex expression.

Understanding function composition is crucial for finding the inverse of composite functions. You essentially need to reverse the order and steps of the original composition.

Domain Restrictions

Domain restrictions are vital when dealing with functions and their inverses. A domain is the set of all possible inputs for a function. Not all functions are naturally invertible across all values of \(x\). This is because inverse functions can bring up issues like division by zero or generating non-real numbers.

• Ensure the function is one-to-one: Each \(x\) maps uniquely to a \(y\).
• With quadratic functions, restrict the domain to where the function is either strictly increasing or decreasing.
• Quadratic equations often have an inverse only for either \(x \geq 0\) or \(x \leq 0\).

By setting domain restrictions, you make sure the function and its inverse are both valid and meaningful. Always examine the range of the original function to set correct restrictions.

Finding Inverses

Finding the inverse of a function involves swapping its input and output values. The goal is to express the original \(x\) as a function of \(y\). For many linear functions, this process is straightforward. Consider these steps:

• Replace \(f(x)\) with \(y\).
• Swap the variables: Set \(x = y\) and \(y = x\).
• Solve the equation for \(y\): Rearrange it to express \(y\) explicitly.

Essentially, the inverse will 'undo' whatever the original function does. For a function \(f\), \(f^{-1}(f(x)) = x\) and \(f(f^{-1}(x)) = x\). This is valuable for scenarios where knowing the original input is essential, such as decoding encoded messages.

Quadratic Functions

Quadratic functions are polynomials with the general form \(f(x) = ax^2 + bx + c\). They shape as parabolas on a graph and can open upwards or downwards. An important aspect of quadratic functions is their symmetry, which can complicate finding inverses.

• Quadratics need domain restrictions to have an inverse; otherwise, they are not one-to-one.
• Solve the equation \(x = y^2 + c\) for \(y\) by isolating and taking the square root: \(y = \sqrt{x - c}\).
• The choice of \(x \geq 0\) or \(x \leq 0\) determines which part of the parabola you focus on.

By restricting the domain and understanding the symmetry of parabolas, you can effectively manage and find the inverses of quadratic functions. This can be practical in physics and other sciences, where quadratic relationships are common.