Lesson Notes By Weeks and Term v5 - Grade 11
Functions – Week 5 focus
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Functions – Week 5 focus
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Subject: Mathematics
Class: Grade 11
Term: 2nd Term
Week: 5
Theme: General lesson support
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This week, we delve deeper into the world of functions, focusing on inverse functions and their graphical representation. Understanding inverse functions is crucial for various applications, from decoding encrypted messages to optimising resource allocation. In South Africa, understanding functions is vital for careers in engineering, finance, and data science, all critical for our nation's development.
Furthermore, the graphical understanding of functions allows us to model real-world phenomena like population growth, financial investments, and even the spread of diseases.
2.1 What is an Inverse Function? An inverse function "undoes" what the original function does. If a function f(x) takes an input x and produces an output y, then the inverse function, denoted as f -1 (y), takes y as input and produces x as output. Mathematically, this means: If f(x) = y, then f -1 (y) = x f -1 (f(x)) = x for all x in the domain of f f(f -1 (x)) = x for all x in the domain of f -1 * Important
Note: Only one-to-one functions have inverses. A one-to-one function is a function where each input has a unique output, and each output has a unique input. Graphically, this can be determined using the horizontal line test: if any horizontal line intersects the graph of the function at most once, then the function is one-to-one. Quadratic functions are not one-to-one over their entire domain, but by restricting the domain, we can create a one-to-one function and find its inverse. 2.2 Finding the Inverse Function Algebraically To find the inverse of a function f(x), follow these steps: Replace f(x) with y. Swap x and y. This is the crucial step where you are essentially "undoing" the original function. Solve for y. This will give you y = f -1 (x). *Replace y with f -1 (x).
Example 1: Linear Function Let f(x) = 2x +
3. Find f -1 (x). y = 2x + 3 x = 2y + 3 x - 3 = 2y y = (x - 3) / 2 f -1 (x) = (x - 3) / 2 Example 2: Quadratic Function (with Restricted Domain) Let f(x) = x 2 - 4, for x ≥
0. Find f -1 (x). y = x 2 - 4 x = y 2 - 4 x + 4 = y 2 y = ±√(x + 4) Since the domain of f(x) is x ≥ 0, the range of f -1 (x) must also be ≥
0. Therefore, we choose the positive square root: f -1 (x) = √(x + 4)
Example 3: Exponential Function Let f(x) = 3 x . Find f -1 (x). y = 3 x x = 3 y To solve for y, we need to use logarithms. The inverse of an exponential function is a logarithmic function. y = log 3 (x) f -1 (x) = log 3 (x) 2.3 Graphical Representation of Inverse Functions The graph of an inverse function is a reflection of the graph of the original function across the line y = x. This is because the x and y coordinates are swapped in the inverse function. If the point (a, b) lies on the graph of f(x), then the point (b, a) lies on the graph of f -1 (x)*. If f(x) has an x-intercept at (c, 0), then f -1 (x) has a y-intercept at (0, c)*. If f(x) has a y-intercept at (0, d), then f -1 (x) has an x-intercept at (d, 0)*. If f(x) has a horizontal asymptote at y = k, then f -1 (x) has a vertical asymptote at x = k. If f(x) has a vertical asymptote at x = j, then f -1 (x) has a horizontal asymptote at y = j. 2.4 Domain and Range of Inverse Functions The domain of f -1 (x) is equal to the range of f(x), and the range of f -1 (x) is equal to the domain of f(x). This is a direct consequence of the swapping of x and y when finding the inverse. Guided Practice (With Solutions)
Question 1: Find the inverse of the function f(x) = -x +
5. Solution: y = -x + 5 x = -y + 5 x - 5 = -y y = -x + 5 f -1 (x) = -x + 5
Commentary: In this case, the inverse function is the same as the original function. This is because f(x) is symmetric about the line y = x.
Question 2: Find the inverse of the function g(x) = x 2 + 1 for x ≥
0. Solution: y = x 2 + 1 x = y 2 + 1 x - 1 = y 2 y = ±√(x - 1) Since x ≥ 0 for g(x), y ≥ 0 for g -1 (x).
Thus: g -1 (x) = √(x - 1)
Commentary: Restricting the domain of g(x) to x ≥ 0 is crucial because it ensures that g(x) is a one-to-one function, allowing us to find a unique inverse. The domain of g -1 (x) is x ≥
1. Question 3: Find the inverse of h(x) = 2 x -
3. Solution: y = 2 x - 3 x = 2 y - 3 x + 3 = 2 y y = log 2 (x + 3) h -1 (x) = log 2 (x + 3)
Commentary: The inverse function is a logarithmic function. The vertical asymptote of the inverse function is at x = -
3. Question 4: Sketch the graph of f(x) = x + 2 and its inverse on the same axes.
Solution: f(x) = x + 2 is a straight line with a y-intercept of 2 and a gradient of
1. Its inverse is found as follows: y = x + 2 x = y + 2 y = x - 2 f -1 (x) = x - 2 f -1 (x) = x - 2 is a straight line with a y-intercept of -2 and a gradient of
1. Sketch both lines on the same set of axes, including the line y = x to demonstrate the reflection.
Commentary: The graph of f -1 (x) is a reflection of f(x) over the line y = x. The x-intercept of f(x) becomes the y-intercept of f -1 (x), and vice-versa. Independent Practice (Questions Only) Find the inverse of f(x) = 5x -
2. Find the inverse of g(x) = (x + 1) /
3. Find the inverse of h(x) = x 2 + 4 for x ≤
0. Find the inverse of k(x) = -√(x - 2) for x ≥ 2. (State domain/range restriction if needed) Find the inverse of m(x) = 4 x +
1. Find the inverse of n(x) = (1/2) x -
5. Sketch the graph of f(x) = x 2 , x>=0 and its inverse on the same axes. Label key points. Sketch the graph of g(x) = 3 x and its inverse on the same axes. Label key points. The function p(x) = x+5 is defined for all x. What is p -1 (10)? q(x) = 2x-
1. Find the value of x for which q(x) = q -1 (x).