Lesson Notes By Weeks and Term v5 - Grade 11
Functions – Week 3 focus
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Functions – Week 3 focus
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Subject: Mathematics
Class: Grade 11
Term: 2nd Term
Week: 3
Theme: General lesson support
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This week, we delve deeper into the fascinating world of functions, specifically focusing on inverse functions and logarithmic functions, and how transformations affect these functions. Understanding these concepts is crucial because functions are the bedrock of mathematical modelling. From predicting population growth to understanding financial investments, functions help us make sense of the world around us. In South Africa, understanding compound interest (an exponential function), for instance, is critical for responsible financial planning, and analyzing trends in unemployment rates (which can be modelled by various functions) helps us understand societal challenges.
2.1 Inverse Functions The inverse of a function, denoted as f⁻¹(x), "undoes" the action of the original function, f(x). If f(a) = b, then f⁻¹(b) = a. For a function to have an inverse, it must be a one-to-one function, meaning that each y-value corresponds to only one x-value (horizontal line test).
Finding the Inverse Algebraically: Replace f(x) with y: Write the function as y = f(x).
Swap x and y: Interchange x and y in the equation.
Solve for y: Solve the new equation for y. This new equation represents the inverse function. Replace y with f⁻¹(x): Write the inverse function using the notation 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 Why it works: Swapping x and y essentially reflects the function across the line y = x. Solving for y isolates the inverse operation.
Example 2: Find the inverse of g(x) = x² (for x ≥ 0). y = x² x = y² y = ±√x Since we restricted the domain of the original function to x ≥ 0, the range of the inverse will be y ≥
0. Therefore, y = √x g⁻¹(x) = √x Graphical Representation: The graph of the inverse function is the reflection of the original function across the line y = x. 2.2 Logarithmic Functions A logarithmic function is the inverse of an exponential function. The logarithmic function y = logₐ(x) asks the question: "To what power must we raise 'a' to get 'x'?" Definition: y = logₐ(x) if and only if x = aʸ, where a > 0, a ≠ 1, and x > 0. 'a' is the base of the logarithm.
Key Properties: Domain: x > 0 (Logarithms are only defined for positive values).
Range: All real numbers.
Asymptote: The y-axis (x = 0) is a vertical asymptote.
Intercept: The x-intercept is (1, 0) because logₐ(1) = 0 for any base 'a'. logₐ(a) = 1 logₐ(aˣ) = x a^(logₐ(x)) = x Example 3: Sketch the graph of y = log₂(x).
Convert to exponential form: x = 2ʸ. Choose y-values and calculate corresponding x-values: y = -2, x = 2⁻² = 1/4 y = -1, x = 2⁻¹ = 1/2 y = 0, x = 2⁰ = 1 y = 1, x = 2¹ = 2 y = 2, x = 2² = 4 Plot the points and draw the curve: The graph will approach the y-axis but never touch it (vertical asymptote), and it will pass through (1, 0) and (2, 1).
Example 4: Simplify log₃(9).
We are asking: "To what power must we raise 3 to get 9?" Since 3² = 9, log₃(9) = 2. 2.3 Relationship between Exponential and Logarithmic Functions As mentioned earlier, logarithmic and exponential functions are inverses of each other.
This means that: If f(x) = aˣ, then f⁻¹(x) = logₐ(x). logₐ(aˣ) = x a^(logₐ(x)) = x Understanding this relationship is crucial for solving exponential and logarithmic equations.
Example 5: Solve for x: 2ˣ = 8 We can rewrite this in logarithmic form: log₂(8) = x. Since 2³ = 8, x = 3. 2.4 Transformations of Functions Transformations alter the graph of a function.
The key transformations are: Vertical Shift: y = f(x) + k (shifts the graph up by k units if k > 0, down by |k| units if k 0, left by |h| units if h 1, compresses vertically by a factor of 'a' if 0 1, stretches horizontally by a factor of 'b' if 0 x = 24 Guided Practice (With Solutions)
Question 1: Find the inverse of f(x) = (x - 2)/
3. Solution: y = (x - 2)/3 x = (y - 2)/3 3x = y - 2 y = 3x + 2 f⁻¹(x) = 3x + 2
Commentary: We followed the standard steps for finding the inverse: swapping x and y and then solving for y.
Question 2: Sketch the graph of y = log₃(x) and identify its asymptote.
Solution: Asymptote: x = 0 (y-axis)
Key points: x = 1/3, y = -1 x = 1, y = 0 x = 3, y = 1 Draw a smooth curve through the points, approaching the y-axis but never touching it.
Commentary: Remember that the graph of y=log₃(x) will have the same general shape as y=log₂(x), just slightly different since the base is different. The key is to identify the vertical asymptote and a few key points to plot.
Question 3: Solve for x: log₄(x) = 1/2 Solution: Convert to exponential form: x = 4^(1/2)
Simplify: x = √4 x = 2
Commentary: Recognizing that 4^(1/2) is the same as the square root of 4 is crucial for solving this problem.
Question 4: Describe the transformation of y = -log₂(x) compared to y = log₂(x).
Solution: The graph of y = log₂(x) has been reflected across the x-axis.
Commentary: The negative sign in front of the function causes a reflection across the x-axis. Independent Practice (Questions Only) Find the inverse of h(x) = √(x + 1) for x ≥ -
1. Sketch the graph of y = log₁₀(x) and state its domain and range.
Solve for x: log₂(3x - 1) =
4. Describe the transformation of y = log₂(x - 2) + 1 compared to y = log₂(x). Find the inverse of f(x) = 5ˣ.
Solve for x: 3^(x+1) = 9 Sketch the graph of y = 2^(x) -1 and y=2^(x-1) Find the equation of the inverse of g(x) = (1/2)x - 3 Evaluate log₅(125) Evaluate 2^(log₂(8))