Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Determine the inverse of a function algebraically and graphically.
• Use the horizontal line test to see if an inverse exists.
• Restrict the domain of a function to create an inverse.
• Sketch graphs of functions and their inverses.
• Apply these concepts to solve problems.

Lesson summary

This week, we delve deeper into the fascinating world of functions and their inverses. Understanding functions and inverses is a fundamental building block for more advanced mathematics, including calculus and other fields. It's not just about manipulating equations; it's about understanding relationships between variables and how to "undo" those relationships.

Think of it like this: a function is like a recipe, taking ingredients (inputs) and creating a dish (output). The inverse function is like figuring out which ingredients were used to create that dish. In South Africa, this understanding can be applied in many practical ways.

Lesson notes

2.1 What is a Function? A function is a relationship between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output.

Think of a function as a machine: you put something in (the input), and it produces something else (the output).

Domain: The set of all possible input values for a function.

Range: The set of all possible output values for a function.

Notation: We usually write a function as f(x), where x is the input and f(x)* is the output. 2.2 What is an Inverse Function? The inverse of a function f(x), denoted as f -1 (x), "undoes" what the function does. In other words, if f(a) = b, then f -1 (b) = a. The domain of f -1 (x) is the range of f(x), and the range of f -1 (x) is the domain of f(x). 2.3 How to Find the Inverse of a Function Algebraically: Replace f(x) with y. Swap x and y. Solve for y. Replace y with f -1 (x).

Example 1: Find the inverse of f(x) = 2x + 3. y = 2x + 3 x = 2y + 3 x - 3 = 2y y = (x - 3) / 2 f -1 (x) = (x - 3) / 2 Example 2: Find the inverse of g(x) = x 2 , x ≥ 0. (Important to note the domain restriction). y = x 2 x = y 2 y = ±√x Since the original domain has x ≥ 0, then the range will always be positive.

Therefore, the domain of the inverse function will need to be restricted to x ≥ 0, and we take the positive solution so the function remains a function. g -1 (x) = √x, x ≥ 0 2.4 Horizontal Line Test: A function has an inverse function if and only if it is a one-to-one function. A function is one-to-one if no horizontal line intersects its graph more than once. This is known as the horizontal line test. If a horizontal line intersects the graph more than once, it means there are two different x-values that produce the same y-value, violating the condition for an inverse to exist. 2.5 Restricting the Domain: If a function is not one-to-one, we can sometimes restrict its domain to make it one-to-one.

Example 3: Consider the function h(x) = x 2 . This is a parabola and fails the horizontal line test.

However, if we restrict the domain to x ≥ 0, the function becomes one-to-one, and its inverse is h -1 (x) = √x, x ≥

0. Similarly, if we restrict the domain to x ≤ 0, the inverse is h -1 (x) = -√x, x ≥ 0. 2.6 Graphing Functions and Their Inverses: The graph of a function and its inverse are reflections of each other in the line y = x. This is because the x and y values are swapped when finding the inverse.

Example 4: Consider f(x) = 2 x and its inverse f -1 (x) = log 2 (x). The exponential function has a horizontal asymptote at y=0, while the logarithmic function has a vertical asymptote at x=

0. Note how the x and y intercepts of f(x) are reversed on the graph of f -1 (x). 2.7 Exponential and Logarithmic functions Exponential and logarithmic functions are inverses of each other. The inverse of f(x) = a x is f -1 (x) = log a x Guided Practice (With Solutions)

Question 1: Find the inverse of f(x) = 3x -

5. Solution: y = 3x - 5 x = 3y - 5 x + 5 = 3y y = (x + 5) / 3 f -1 (x) = (x + 5) / 3

Commentary: This is a straightforward linear function. The process involves swapping x and y and then isolating y.

Question 2: Determine if g(x) = x 2 + 2 has an inverse without restricting its domain.

Solution: The graph of g(x) = x 2 + 2 is a parabola shifted upward by 2 units. It fails the horizontal line test because a horizontal line drawn above y = 2 will intersect the graph at two points.

Therefore, g(x) does not have an inverse unless we restrict its domain.

Commentary: Applying the horizontal line test is crucial for determining if an inverse exists. Recognizing the shape of common functions (like parabolas) can speed up this process.

Question 3: Restrict the domain of h(x) = x 2 - 4x + 3 such that its inverse exists. Then, find the inverse.

Solution: First, complete the square: h(x) = (x - 2) 2 -

1. This is a parabola with vertex at (2, -1). To make it one-to-one, we can restrict the domain to x ≥ 2 or x ≤

2. Let's choose x ≥ 2. y = (x - 2) 2 - 1 x = (y - 2) 2 - 1 x + 1 = (y - 2) 2 y - 2 = ±√(x + 1) Because we restricted the function to x ≥ 2, we need the positive branch of the square root. y = 2 + √(x + 1) h -1 (x) = 2 + √(x + 1), x ≥ -1

Commentary: Completing the square helps identify the vertex of the parabola, which aids in domain restriction. Remember that the domain of the inverse will be the range of the original function, hence x ≥ -

1. Question 4: Sketch the graph of f(x) = 2 x and its inverse.

Solution: f(x) = 2 x * is an exponential function passing through (0, 1) and (1, 2) and has a horizontal asymptote at y =

0. Its inverse is f -1 (x) = log 2 (x)*, which passes through (1, 0) and (2, 1) and has a vertical asymptote at x =

0. The two graphs are reflections of each other over the line y = x.

Commentary: Pay attention to key points (intercepts, asymptotes) when sketching. Remember that the coordinates switch when finding the inverse. Independent Practice (Questions Only) Find the inverse of f(x) = (x + 1) / (x - 2).