Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Determine and sketch inverse functions.
• Solve exponential and logarithmic equations.
• Calculate terms and sums of sequences and series.
• Apply financial maths formulas to solve problems.
• Distinguish between different financial scenarios.

Lesson summary

This week focuses on revising crucial topics frequently assessed in the final Grade 12 Mathematics examination. We will consolidate our understanding of functions (specifically inverse functions, logarithms, and exponential functions), sequences and series (arithmetic and geometric), and financial mathematics (including annuities and sinking funds). These topics are fundamental building blocks for further studies in mathematics, science, engineering, and finance. They also equip us with the analytical skills needed to make informed decisions in our daily lives, from understanding interest rates on loans to projecting investment growth.

Lesson notes

A. Inverse Functions The inverse of a function, denoted by f⁻¹(x), "undoes" the action of the original function, f(x). If f(a) = b, then f⁻¹(b) = a.

Finding the Inverse: Replace f(x) with y. Interchange x and y. Solve for y. Replace y with f⁻¹(x).

Domain and Range: The domain of f(x) becomes the range of f⁻¹(x), and the range of f(x) becomes the domain of f⁻¹(x).

One-to-One Functions: Only one-to-one functions (functions that pass the horizontal line test) have inverses that are also functions. If a function is not one-to-one, we may need to restrict its domain to create an inverse.

Graphs: The graph of f⁻¹(x) is a reflection of the graph of f(x) about 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: Find the inverse of g(x) = x² (x ≥ 0). Note the domain restriction. y = x² x = y² y = ±√x Because the domain of g(x) is restricted to x ≥ 0, we only consider the positive square root: y = √x g⁻¹(x) = √x

B. Exponential and Logarithmic Functions Exponential Functions: Functions of the form f(x) = aˣ, where a > 0 and a ≠

1. Logarithmic Functions: The inverse of an exponential function. If aˣ = y, then logₐ(y) = x.

Key Properties of Logarithms: logₐ(1) = 0 logₐ(a) = 1 logₐ(xy) = logₐ(x) + logₐ(y) logₐ(x/y) = logₐ(x) - logₐ(y) logₐ(xⁿ) = n logₐ(x)

Change of Base Formula: logₐ(x) = logₓ(x) / logₓ(a) This allows you to use a calculator to find logarithms to any base.

Example 3: Solve for x: 2ˣ = 8 2ˣ = 2³ x = 3 Example 4: Solve for x: 3^(x+1) = 27 3^(x+1) = 3³ x + 1 = 3 x = 2 Example 5: Solve for x: log₂(x) = 3 2³ = x x = 8 Example 6: Solve for x: log₃(x + 2) + log₃(x - 2) = 1 log₃[(x + 2)(x - 2)] = 1 log₃(x² - 4) = 1 3¹ = x² - 4 x² = 7 x = ±√7 Since we cannot take the logarithm of a negative number, we reject x = -√

7. Therefore, x = √

7. C.

Sequences and Series Arithmetic Sequence: A sequence where the difference between consecutive terms is constant (the common difference, d). The general term is Tₙ = a + (n - 1)d, where a is the first term. The sum of the first n terms is Sₙ = n/2 [2a + (n - 1)d] or Sₙ = n/2 [a + l], where l is the last term.

Geometric Sequence: A sequence where the ratio between consecutive terms is constant (the common ratio, r). The general term is Tₙ = ar^(n-1), where a is the first term. The sum of the first n terms is Sₙ = a(rⁿ - 1) / (r - 1) (r ≠ 1). The sum to infinity exists only when |r| < 1, and is given by S∞ = a / (1 - r).

Example 7: Consider the arithmetic sequence: 2, 5, 8, 11, ... a = 2, d = 3 Find T₁₀: T₁₀ = 2 + (10 - 1)(3) = 2 + 27 = 29 Find S₁₀: S₁₀ = 10/2 [2(2) + (10 - 1)(3)] = 5[4 + 27] = 5(31) = 155 Example 8: Consider the geometric sequence: 3, 6, 12, 24, ... a = 3, r = 2 Find T₇: T₇ = 3(2)^(7-1) = 3(2⁶) = 3(64) = 192 Find S₇: S₇ = 3(2⁷ - 1) / (2 - 1) = 3(128 - 1) / 1 = 3(127) = 381 Example 9: Consider the geometric sequence: 8, 4, 2, 1,... a = 8, r = 1/2 Find S∞: S∞ = 8 / (1 - 1/2) = 8 / (1/2) = 16

D. Financial Mathematics Simple Interest: A = P(1 + in), where A is the accumulated amount, P is the principal, i is the interest rate per period, and n is the number of periods.

Compound Interest: A = P(1 + i)ⁿ Annuities: A series of equal payments made at regular intervals.

Future Value Annuity: FV = x [((1 + i)ⁿ - 1) / i], where FV is the future value, x is the regular payment, i is the interest rate per period, and n is the number of periods. This is used when saving towards a goal.

Present Value Annuity: PV = x [(1 - (1 + i)⁻ⁿ) / i], where PV is the present value, x is the regular payment, i is the interest rate per period, and n is the number of periods. This is used for loans or investments where you receive regular payments.

Sinking Funds: A type of future value annuity where you are saving to accumulate a specific amount.

Example 10: You invest R5000 at an interest rate of 8% per annum compounded annually for 5 years. Calculate the accumulated amount. A = 5000(1 + 0.08)⁵ = 5000(1.08)⁵ = R7346.64 Example 11: You want to buy a car that costs R200,

0

0

0. You take out a loan at an interest rate of 12% per annum compounded monthly, and you plan to repay the loan over 5 years. What will be your monthly payment? PV = 200000, i = 0.12/12 = 0.01, n = 5 12 = 60 200000 = x [(1 - (1 + 0.01)⁻⁶⁰) / 0.01] x = 200000 0.01 / (1 - (1.01)⁻⁶⁰) = R4448.89 Example 12: You want to save R100,000 in 10 years. You can earn an interest rate of 6% per annum compounded monthly. What monthly payment do you need to make? FV = 100000, i = 0.06/12 = 0.005, n = 10 12 = 120 100000 = x [((1 + 0.005)¹²⁰ - 1) / 0.005] x = 100000 0.005 / ((1.005)¹²⁰ - 1) = R614.47 Guided Practice (With Solutions)

Question 1: Determine 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: This is a straightforward linear function. Remember to interchange x and y and solve for y.

Question 2: Solve for x: 5^(2x - 1) = 125.