Question

Match the function with its inverse. Function a) \(y=2 x+5\) b) \(y=\frac{1}{2} x-4\) c) \(y=6-3 x\) d) \(y=x^{2}-12, x \geq 0\) e) \(y=\frac{1}{2}(x+1)^{2}, x \leq-1\) Inverse \(\mathbf{A} \quad y=\sqrt{x+12}\) \(\mathbf{B} \quad y=\frac{6-x}{3}\) \(\mathbf{c} \quad y=2 x+8\) \(\mathbf{D} \quad y=-\sqrt{2 x}-1\) \(\mathbf{E} \quad y=\frac{x-5}{2}\)

Short answer

a) E, b) C, c) B, d) A, e) D

Step-by-step solution

- Understanding the concept of inverse functions

The inverse function reverses the operation of the original function. For instance, if the function is f(x) and its inverse is g(y), then if we apply g on the result of f, we should get back our original value: g(f(x)) = x.

- Finding the inverse of each function

For each function given, we'll find its inverse by solving for x in terms of y and then check which provided inverse matches.

- Inverse of function a: y = 2x + 5

Start by solving for x: \[y = 2x + 5\] 1. Subtract 5 from both sides: \[y - 5 = 2x\] 2. Divide by 2: \[x = \frac{y - 5}{2}\] Therefore, the inverse is \(y = \frac{x - 5}{2}\), which matches option E.

- Inverse of function b: y = \frac{1}{2}x - 4

Start by solving for x: \[y = \frac{1}{2}x - 4\] 1. Add 4 to both sides: \[y + 4 = \frac{1}{2}x\] 2. Multiply by 2: \[x = 2(y + 4)\] Simplify: \[x = 2y + 8\] Thus, the inverse is \(y = 2x + 8\), which matches option C.

- Inverse of function c: y = 6 - 3x

Start by solving for x: \[y = 6 - 3x\] 1. Subtract 6 from both sides: \[y - 6 = -3x\] 2. Divide by -3: \[x = \frac{6 - y}{3}\] Therefore, the inverse is \(y = \frac{6 - x}{3}\), which matches option B.

- Inverse of function d: y = x^2 - 12, x ≥ 0

Start by solving for x: \[y = x^2 - 12\] 1. Add 12 to both sides: \[y + 12 = x^2\] 2. Take the square root of both sides: \[x = \sqrt{y + 12}\] Therefore, the inverse is \(y = \sqrt{x + 12}\), which matches option A.

- Inverse of function e: y = \frac{1}{2}(x + 1)^2, x ≤ -1

Start by solving for x: \[y = \frac{1}{2}(x + 1)^2\] 1. Multiply by 2: \[2y = (x + 1)^2\] 2. Take the square root of both sides (note that x ≤ -1 implies we take the negative square root): \[x + 1 = -\sqrt{2y}\] 3. Subtract 1 from both sides: \[x = -\sqrt{2y} - 1\] Therefore, the inverse is \(y = -\sqrt{2x} - 1\), which matches option D.

Key concepts

Solving Equations

Solving equations is a fundamental skill in understanding inverse functions. When we solve equations, we isolate the variable on one side of the equation. For example, if we have the equation \(y = 2x + 5\), we solve for \(x\) by performing the following steps:

• First, subtract 5 from both sides to get \(y - 5 = 2x\).
• Then, divide both sides by 2, which gives us \(x = \frac{y-5}{2} \).

This approach helps in finding the inverse of the function, as we essentially reverse the operations performed on the input variable. Remember, each arithmetic operation has an inverse operation:

• Addition and subtraction are inverses
• Multiplication and division are inverses

Practicing these steps will make it easier to master more complex functions.

Function Operations

Understanding how to perform operations with functions is critical when dealing with inverses. Function operations include addition, subtraction, multiplication, and division of functions. For instance, when you add two functions, you combine their outputs:

• If \(f(x) = x + 2\) and \(g(x) = 3x\), then \((f + g)(x) = (x + 2) + 3x = 4x + 2\)

Function composition is another important operation, especially relevant to inverse functions. When we compose two functions \(f(g(x))\) or \(g(f(x))\), we apply one function to the result of another. If these functions are inverses, their compositions will undo each other:

• For example, if \(f(x) = 2x + 5\) and \(f^{-1}(x) = \frac{x-5}{2}\), then \(f(f^{-1}(x)) = x\).

This principle is very powerful and is the basis for verifying that two functions are indeed inverses.

Precalculus Problems

Precalculus problems often involve concepts from algebra and trigonometry, especially when exploring the properties and inverses of various functions. Solving such problems helps build a strong foundation for calculus. One common type of problem is matching functions with their inverses.
For example, given a function \(y = 2x + 5\), you might be asked to find its inverse. We solve by isolating \(x\) on one side:

• Subtract 5: \(y - 5 = 2x\)
• Divide by 2: \(x = \frac{y-5}{2}\)

Matching this with the provided choices helps reinforce understanding. By practicing multiple problems, students become proficient in manipulations needed for more advanced concepts.

Function Transformations

Function transformations involve shifting, stretching, or reflecting a function on a graph. These transformations help in visualizing the effect of changes to the function's equation. Understanding inverses also involves transformations, as we often reflect a function over the line \(y=x\) to find its inverse.
Consider the function \(y = x^2 - 12\). To find its inverse, we solve:

• Add 12: \(y + 12 = x^2\)
• Take the square root: \(x = \sqrt{y + 12}\) (since \(x \geq 0\))

This inverse transformation shows how inputs and outputs swap places and how graph reflections work. Function transformations are visually powerful tools for understanding inverses and other function properties.