Differentiation: Definition and Fundamental Properties: Unit Diagnostic (Calculus AB Unit 2)

What this is for. Use this diagnostic to find which topics of Unit 2, Differentiation: Definition and Fundamental Properties, to revisit. There is one question per topic, and two for Topics 2.2, 2.4 and 2.7. These are original Marlbridge practice questions, not past exam questions. They are not calibrated, and your result is not a predicted score.

Rules. No calculator; about 30 minutes. Angles are in radians. Answer everything before opening any answer. The unit is shared by Calculus AB and Calculus BC, so no question here is BC only.

Question 1 (multiple choice · 2.1)

Let f(x) = x² + 3x. For h ≠ 0, the difference quotient (f(1 + h) − f(1))/h simplifies to h + 5. Which statement is true?

(A) The average rate of change of f over [1, 4] is 8, and f′(1) = 5.
(B) The average rate of change of f over [1, 4] is 8, and f′(1) = 6.
(C) The average rate of change of f over [1, 4] is 24, and f′(1) = 5.
(D) f′(1) cannot be found, because the quotient is 0/0 at h = 0.

Answer and explanation

Answer: (A). The interval [1, 4] is [1, 1 + h] with h = 3, so the average rate is 3 + 5 = 8. (Check: (28 − 4)/(4 − 1) = 8.) The derivative is the limit as h → 0: f′(1) = 0 + 5 = 5.

(B) puts h = 1 instead of taking the limit.
(C) forgets to divide the change in f, 24, by the change in x, 3.
(D) A limit never uses h = 0 itself.

If you missed this: Topic 2.1 study guide.

Question 2 (multiple choice · 2.2)

The line tangent to the graph of g at x = −2 is y = 3x + 7. Which statement must be true?

(A) g(−2) = 7 and g′(−2) = 3
(B) g(−2) = 3 and g′(−2) = 1
(C) g(−2) = 1 and g′(−2) = 3
(D) g′(x) = 3 for every x

Answer and explanation

Answer: (C). The tangent line touches the graph at x = −2, so g(−2) = 3(−2) + 7 = 1. Its slope is the derivative there: g′(−2) = 3.

(A) uses the y-intercept, the line’s height at x = 0.
(B) swaps the value and the slope.
(D) The tangent line gives the slope at one point only.

If you missed this: Topic 2.2 study guide.

Question 3 (short answer · 2.2)

Let f(x) = 1/(2x − 1).

(a) Use the definition of the derivative to find f′(x). (b) Write an equation of the line tangent to the graph of f at x = 1.

Worked answer

(a) Combine the fractions in f(x + h) − f(x):

1/(2x + 2h − 1) − 1/(2x − 1) = [(2x − 1) − (2x + 2h − 1)] / [(2x + 2h − 1)(2x − 1)] = −2h / [(2x + 2h − 1)(2x − 1)].

Divide by h (h ≠ 0): −2 / [(2x + 2h − 1)(2x − 1)]. Let h → 0: f′(x) = −2/(2x − 1)², for x ≠ 1/2.

(b) Point: f(1) = 1/1 = 1. Slope: f′(1) = −2/1 = −2. Tangent line: y − 1 = −2(x − 1), or y = −2x + 3.

If you missed this: Worked examples 2 and 3 in the Topic 2.2 study guide.

Question 4 (multiple choice · 2.3)

The height H(t) of a fictional sunflower, in centimetres, is measured t days after the start of an experiment. H is differentiable. The data are invented.

t (days)	0	5	8	12	20
H(t) (cm)	50	62	71	77	81

What is the best estimate of H′(10) from the table?

(A) 1.55 cm per day
(B) 1.5 cm per day
(C) 3 cm per day
(D) 6 cm per day

Answer and explanation

Answer: (B). Use the shortest interval in the table that contains t = 10, which is [8, 12]: (H(12) − H(8))/(12 − 8) = (77 − 71)/4 = 1.5 cm per day.

(A) uses [0, 20], which is far too wide.
(C) uses [5, 8], which does not contain t = 10.
(D) is the change in height, 6 cm, not divided by the 4 days.

If you missed this: Topic 2.3 study guide.

Question 5 (multiple choice · 2.4)

A function f is not continuous at x = 5. Which statement must be true?

(A) f(5) is undefined.
(B) lim (x → 5) f(x) does not exist.
(C) The graph of f has a vertical asymptote at x = 5.
(D) f′(5) does not exist.

Answer and explanation

Answer: (D). Differentiable at 5 would mean continuous at 5. f is not continuous there, so it is not differentiable there.

(A) and (B) can cause a discontinuity, but neither must happen: f(5) and the limit can both exist and differ.
(C) A jump or a hole is also a discontinuity.

If you missed this: Topic 2.4 study guide.

Question 6 (short answer · 2.4)

Let f(x) = x² + 1 for x ≤ 1, and f(x) = 4√x − 2 for x > 1.

(a) Show that f is continuous at x = 1. (b) Use one-sided limits of the difference quotient to decide whether f is differentiable at x = 1. (c) If the second piece were 4√x instead, would f be differentiable at x = 1? Give a reason.

Worked answer

(a) f(1) = 1 + 1 = 2. Left-hand limit: 1² + 1 = 2. Right-hand limit: 4√1 − 2 = 2. The limit is 2 = f(1), so f is continuous at x = 1.

(b) From the left: ((1 + h)² + 1 − 2)/h = (2h + h²)/h = 2 + h → 2. From the right: (4√(1 + h) − 2 − 2)/h = 4(√(1 + h) − 1)/h. Multiply by the conjugate: 4/(√(1 + h) + 1) → 4/2 = 2. Both one-sided limits equal 2, so f is differentiable at x = 1, with f′(1) = 2.

(c) No. The right-hand limit would be 4, not f(1) = 2, so f would not be continuous at 1, and so not differentiable there, even though the slopes still match.

If you missed this: Worked example 2 in the Topic 2.4 study guide.

Question 7 (multiple choice · 2.5)

Let f(x) = 3/∛x for x > 0. What is f′(8)?

(A) −1/4
(B) −1/2
(C) −1/16
(D) 36

Answer and explanation

Answer: (C). Rewrite f(x) = 3x^(−1/3). Then f′(x) = 3 × (−1/3)x^(−4/3) = −x^(−4/3). Since 8^(1/3) = 2, 8^(4/3) = 16, so f′(8) = −1/16.

(A) subtracts 1/3 instead of 1 from the exponent, giving −x^(−2/3).
(B) brings the power down but forgets to subtract 1.
(D) applies the power rule to the denominator alone, as if the derivative of 3/u were 3/u′.

If you missed this: Topic 2.5 study guide.

Question 8 (multiple choice · 2.6)

The line tangent to y = x³ − 2x² + kx at x = 2 is parallel to the line y = 7x. What is the constant k?

(A) −1
(B) 3
(C) 7/2
(D) 7

Answer and explanation

Answer: (B). Parallel lines have equal slopes, so dy/dx = 7 at x = 2. Term by term, dy/dx = 3x² − 4x + k. At x = 2: 12 − 8 + k = 7, so k = 3.

(A) differentiates −2x² as −2x.
(C) sets the y-value, 8 − 8 + 2k, equal to 7 instead of the slope.
(D) assumes the slope of kx alone must be 7.

If you missed this: Topic 2.6 study guide.

Question 9 (multiple choice · 2.7)

Let f(x) = 4 ln x − x² for x > 0. At which x does the graph of f have a horizontal tangent line?

(A) x = −√2 and x = √2
(B) x = 2
(C) x = e
(D) x = √2

Answer and explanation

Answer: (D). f′(x) = 4/x − 2x. Set it equal to 0: 4/x = 2x, so x² = 2. Only x = √2 is in the domain x > 0.

(A) forgets that ln x needs x > 0.
(B) solves 4/x = 2 instead of 4/x = 2x.
(C) does not solve f′(x) = 0.

If you missed this: Topic 2.7 study guide.

Question 10 (multiple choice · 2.7)

What is lim (x → 4) (ln x − ln 4)/(x − 4)?

(A) 1/4
(B) ln 4
(C) 0
(D) It does not exist, because substitution gives 0/0.

Answer and explanation

Answer: (A). This is the x-form of the definition, lim (x → a) (f(x) − f(a))/(x − a), with f(x) = ln x and a = 4. Since d/dx ln x = 1/x, the limit is f′(4) = 1/4.

(B) is f(4), the value, not the slope.
(C) assumes the top tending to 0 makes the whole fraction 0.
(D) 0/0 means “find another method”, here recognising a derivative.

If you missed this: Worked example 3 in the Topic 2.7 study guide.

Question 11 (multiple choice · 2.8)

Let y = √x · eˣ. What is dy/dx at x = 1?

(A) e/2
(B) e
(C) 3e/2
(D) 2e

Answer and explanation

Answer: (C). Product rule: dy/dx = (1/(2√x))eˣ + √x·eˣ. At x = 1: e/2 + e = 3e/2.

(A) multiplies the two derivatives, (1/2) × e.
(B) keeps only the second term of the product rule.
(D) uses x^(−1/2) as the derivative of √x, dropping the factor 1/2.

If you missed this: Topic 2.8 study guide.

Question 12 (multiple choice · 2.9)

Let f(x) = (cos x)/x for x ≠ 0. What is f′(π/2)?

(A) 2/π
(B) −2/π
(C) −1
(D) −π/2

Answer and explanation

Answer: (B). Quotient rule: f′(x) = (x(−sin x) − cos x · 1)/x². At x = π/2: (−π/2 − 0)/(π²/4) = −2/π.

(A) reverses the order of the top: cos x · 1 − x(−sin x).
(C) divides the derivatives, −sin x / 1.
(D) forgets to divide by x² = π²/4.

If you missed this: Topic 2.9 study guide.

Question 13 (short answer · 2.10)

Let f(x) = sec x + tan x for −π/2 < x < π/2.

(a) Show that f′(x) = sec x (sec x + tan x). (b) Write an equation of the line tangent to the graph of f at x = 0. (c) Explain why f′(x) > 0 for every x in the interval. (Hint: write f(x) as one fraction.)

Worked answer

(a) d/dx sec x = sec x tan x and d/dx tan x = sec²x. So f′(x) = sec x tan x + sec²x = sec x (tan x + sec x).

(b) f(0) = 1 + 0 = 1 and f′(0) = 1 × (0 + 1) = 1. Tangent line: y = x + 1.

(c) On (−π/2, π/2), cos x > 0, so sec x > 0. Also sec x + tan x = (1 + sin x)/cos x, and sin x > −1 here, so this factor is positive too. A product of two positive factors is positive.

If you missed this: Topic 2.10 study guide.

Your next step

Topic	Question(s)	If you missed it, read
2.1 Average and instantaneous rates	1	Guide 2.1
2.2 Derivative function, notation, tangent lines	2, 3	Guide 2.2
2.3 Estimating derivatives	4	Guide 2.3
2.4 Differentiability and continuity	5, 6	Guide 2.4
2.5 Power rule	7	Guide 2.5
2.6 Constant, sum, difference rules	8	Guide 2.6
2.7 sin x, cos x, eˣ, ln x	9, 10	Guide 2.7
2.8 Product rule	11	Guide 2.8
2.9 Quotient rule	12	Guide 2.9
2.10 tan, cot, sec, csc	13	Guide 2.10

How to use your result

Mark each topic secure, shaky (unsure or a slip) or gap (wrong).
Fix gaps in Topics 2.1, 2.2 and 2.5 first; every later rule builds on them.
For a gap, read the guide, then the practice set.
Then try the Unit 2 mixed review, where each question combines topics.