Unit notes and practice questions
Everything here is for unit 2 of the Grade 12 Mathematics textbook: what the chapter covers, the short notes, practice questions with their answers, and the lesson video where one exists.
The cards filed under this chapter on the Grade 12 Mathematics page. Open one at a time.
When you use it
Use to calculate the overall rate of change or the slope of the secant line between two endpoints on an interval [a, b].
Watch out
Do not divide f(b) - f(a) by a - b. The order of points in the numerator and denominator must match.
Drafted from Grade 12 Mathematics, pages 61-158, then checked twice before it went up
When you use it
Use when distinguishing between the rate of change across an interval and the rate of change at an exact point.
Watch out
Do not compute the average rate of change formula over an interval when the problem asks for the instantaneous rate at an exact point. An instantaneous rate requires taking the limit as h approaches zero.
Drafted from Grade 12 Mathematics, pages 61-158, then checked twice before it went up
| Slope Value | Graph Direction | Function Behavior |
|---|---|---|
| Positive | Slopes upward from left to right | Increasing |
| Negative | Slopes downward from left to right | Decreasing |
| Zero | Horizontal line | Constant |
| Undefined | Vertical line | Undefined |
When you use it
Use to interpret what the sign of a slope or derivative says about the graph and behavior of a function.
Drafted from Grade 12 Mathematics, pages 61-158, then checked twice before it went up
When you use it
Use to find the exact rate of change or the slope of the tangent line of a function at a single specific point x_0.
Watch out
Do not substitute h = 0 before simplifying the expression, as this produces an undefined division by zero.
Drafted from Grade 12 Mathematics, pages 61-158, then checked twice before it went up
10 multiple-choice questions written against this chapter, each with its answer and the reason. Try one before you open the answer.
Question 1
Find the rate of change for the pair of points (0, 2) and (2, 2).
Answer: B 0
Why
The rate of change is the change in y divided by the change in x. Here the change in y is 2 - 2 = 0, and the change in x is 2 - 0 = 2. So the rate is 0 / 2 = 0. When y does not change, the rate of change is zero.
Question 2
Which pair of points has a negative rate of change?
Answer: A (2, 3) and (3, 1)
Why
For (2, 3) and (3, 1), the change in y is 1 - 3 = -2 and the change in x is 3 - 2 = 1. The rate is -2 / 1 = -2, which is negative. The y-value goes down as x goes up. The other pairs give positive, zero, or undefined rates.
Question 3
A candle in Bahir Dar has a starting length of 10 cm. Thirty minutes after it is lit, the length is 7 cm. What is the average rate of change of the length of the candle?
Answer: A -0.1 cm per minute
Why
The change in length is 7 - 10 = -3 cm. The change in time is 30 minutes. The average rate of change is -3 / 30 = -0.1 cm per minute. The minus sign shows the candle is getting shorter.
Question 4
Find the average rate of change of f(x) = 3x - 2 on the interval [1, 5].
Answer: B 3
Why
First find the function values. f(5) = 3(5) - 2 = 13 and f(1) = 3(1) - 2 = 1. The average rate of change is (f(5) - f(1)) / (5 - 1) = (13 - 1) / 4 = 12 / 4 = 3.
Question 5
For the function f(x) = x2, the gradient of the tangent line at any point P(a, a2) is 2a. What is the gradient of the tangent line at P(4, 16)?
Answer: B 8
Why
The point P(4, 16) has a = 4. Using the formula 2a, the gradient is 2(4) = 8. You can also say the gradient is found by doubling the x-coordinate.
Question 6
Use the chain rule to find f'(1) when f(x) = (2x + 1)3.
Answer: C 54
Why
Use the chain rule: the derivative of the outside power gives 3(2x + 1)2, and the derivative of the inside (2x + 1) is 2. So f'(x) = 3(2x + 1)2 * 2 = 6(2x + 1)2. At x = 1, 2x + 1 = 3, so f'(1) = 6(3)2 = 6(9) = 54.
Question 7
Suppose f'(x) < 0 for all x in an interval I. What can you say about f on I?
Answer: B f is decreasing on I
Why
The increasing and decreasing test says that if f'(x) < 0 for all x in the interior of an interval I, then f is decreasing on I. A negative derivative means the graph falls as x moves to the right.
Question 8
Evaluate the instantaneous rate of change of f(x) = 6x2 - 3 at x = 1.
Answer: B 12
Why
Write f(1 + h) - f(1) = [6(1 + h)2 - 3] - [6(1)2 - 3] = 12h + 6h2. Divide by h to get 12 + 6h. As h gets closer and closer to zero, this approaches 12. So the instantaneous rate of change at x = 1 is 12.
Question 9
Use the product rule to find f'(0) when f(x) = (2x + 1)(x2 + x - 5).
Answer: A -9
Why
The product rule says (uv)' = u'v + uv'. Let u = 2x + 1, so u' = 2, and let v = x2 + x - 5, so v' = 2x + 1. Then f'(x) = 2(x2 + x - 5) + (2x + 1)(2x + 1). At x = 0, f'(0) = 2(-5) + (1)(1) = -10 + 1 = -9.
Question 10
The biomass of a bacterial culture in milligrams, as a function of time t in minutes, is f(t) = 71t3 + 63. What is the instantaneous rate of growth of the culture when t = 2 minutes?
Answer: B 852 mg/min
Why
The derivative of 71t3 + 63 is f'(t) = 213t2. At t = 2, f'(2) = 213(2)2 = 213(4) = 852. The rate is positive, so it is a rate of growth. The answer is 852 mg/min.
The app holds 11 questions on this unit, marked as you go. These are the first 10.
A lesson on this chapter from YouTube. It loads only when you press play.
Introduction to definite integrals · Watch on YouTube
Unit 2 runs from page 61 to page 158 of the Ministry of Education edition.
Ministry of Education edition, English, 426 pages
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