Lesson Notes By Weeks and Term v5 - Grade 11
Functions – Week 4 focus
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Functions – Week 4 focus
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Subject: Mathematics
Class: Grade 11
Term: 2nd Term
Week: 4
Theme: General lesson support
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This week, we delve deeper into the fascinating world of functions, building upon our previous understanding. Functions are fundamental to mathematics and provide a powerful tool for modelling real-world relationships. Understanding functions allows us to predict, analyze, and interpret patterns in diverse scenarios, from tracking the growth of a business to understanding the spread of information. In South Africa, analyzing trends in economic growth, population dynamics, or environmental changes relies heavily on the concepts we'll be covering this week. Specifically, we'll be focusing on inverse functions and how to graph them.
2.1 Inverse Functions Definition: The inverse of a function, denoted as f -1 (x), "undoes" the action of the original function f(x). In other words, if f(a) = b, then f -1 (b) = a. A function must be one-to-one (pass both the vertical and horizontal line tests) to have an inverse function. If a function is not one-to-one, we can sometimes restrict its domain to create a one-to-one function and then find the inverse.
Finding the Inverse Algebraically: Replace f(x) with y. Swap x and y. Solve for y in terms of x. Replace y with f -1 (x).
Graphical Representation: The graph of f -1 (x) is the reflection of the graph of f(x) across the line y = 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: Consider g(x) = x 2 . This is a parabola and doesn't pass the horizontal line test. To find an inverse, we need to restrict the domain. Let's restrict the domain to x ≥
0. Now: y = x 2 x = y 2 y = ±√x Since we restricted the domain to x ≥ 0 for g(x), we have y ≥
0. Therefore, we take the positive square root: y = √x. g -1 (x) = √x 2.2 Exponential Functions Definition: An exponential function is a function of the form f(x) = a x , where a is a constant called the base and a > 0 and a ≠
1. A horizontal asymptote is created by adding the constant q f(x) = a x + q.
Key Features: Intercepts: The y-intercept is found by setting x = 0, so y = a 0 = 1 (unless vertically shifted by adding q). There is no x-intercept when q =
0. Asymptotes: The horizontal asymptote is y = 0 for f(x) = a x , and y = q for f(x) = a x + q.
Domain: The domain is all real numbers, x ∈ ℝ.
Range: The range depends on the base a and the value of q. If a > 1 (exponential growth), the range of f(x) = a x is y > 0 and the range of f(x) = a x + q is y > q. If 0 x is y > 0 and the range of f(x) = a x + q is y > q.
Effects of a and q: a: Determines whether the function is increasing (a > 1) or decreasing (0 x + 1. a = 2 (growth function) q = 1 (vertical shift upwards by 1 unit) y-intercept: f(0) = 2 0 + 1 = 1 + 1 =
2. Horizontal asymptote: y = 1 Domain: x ∈ ℝ Range: y > 1 Example 4: Analyze the function g(x) = (1/2) x - 2. a = 1/2 (decay function) q = -2 (vertical shift downwards by 2 units) y-intercept: g(0) = (1/2) 0 - 2 = 1 - 2 = -
1. Horizontal asymptote: y = -2 Domain: x ∈ ℝ Range: y > -2 2.3 Logarithmic Functions Definition: The logarithmic function is the inverse of the exponential function. If y = a x , then x = log a (y).
The logarithm answers the question: "To what power must we raise a to get y?".
Key Relationship: a log a (x) = x and log a (a x ) = x.
Example 5: Convert the exponential equation 3 2 = 9 to logarithmic form. log 3 (9) = 2 Example 6: Convert the logarithmic equation log 2 (8) = 3 to exponential form. 2 3 = 8 2.4 Solving Basic Exponential Equations Strategy: Express both sides of the equation with the same base. If a x = a y , then x = y.
Example 7: Solve for x: 2 x = 8. 2 x = 2 3 Therefore, x = 3 Example 8: Solve for x: 3 x+1 = 27 3 x+1 = 3 3 Therefore, x + 1 = 3 x = 2 Guided Practice (With Solutions)
Question 1: Find the inverse of h(x) = (3x - 1) /
2. Solution: y = (3x - 1) / 2 x = (3y - 1) / 2 2x = 3y - 1 2x + 1 = 3y y = (2x + 1) / 3 h -1 (x) = (2x + 1) / 3
Commentary: The key is to isolate y after swapping x and y. Remember to follow the order of operations carefully.
Question 2: Sketch the graph of y = 3 x -
2. Identify the y-intercept and the asymptote.
Solution: This is an exponential growth function with a = 3 and q = -
2. The y-intercept is found by setting x = 0: y = 3 0 - 2 = 1 - 2 = -
1. The horizontal asymptote is y = -
2. The graph will approach the line y = -2 as x approaches negative infinity, and will increase rapidly as x increases.
Commentary: Focus on recognizing the base (a) to determine growth or decay and the vertical shift (q) which determines the asymptote.
Question 3: Solve for x: 5 x =
1
2
5. Solution: 5 x = 5 3 Therefore, x = 3
Commentary: Recognizing that 125 is a power of 5 is crucial. Sometimes, you need to rewrite both sides as powers of a common base.
Question 4: Convert the equation log 4 (16) = 2 into exponential form.
Solution: 4 2 = 16
Commentary: Ensure you understand the relationship between logarithms and exponents. The base of the logarithm becomes the base of the exponent.
Question 5: Determine the equation of the asymptote of f(x) = 2 x +
5. Solution: The asymptote is at y =
5. Commentary: The asymptote of the function f(x) = a x + q is always y = q. Independent Practice (Questions Only) Find the inverse of f(x) = 5x -
2. Find the inverse of g(x) = √(x + 1). Restrict the domain of g(x) appropriately. Sketch the graph of y = (1/3) x +
1. Identify the y-intercept and the asymptote. Sketch the graph of y = 4 x -
3. Identify the y-intercept and the asymptote.
Solve for x: 4 x =
6
4. Solve for x: (1/2) x =
4. Solve for x: 2 x-1 =
1
6. Convert the equation log 5 (25) = 2 into exponential form.