NBT Mathematics: Differential Calculus — Questions and Answers
Question 1: If f(x) = 3x⁴ − 2x² + 5x − 1, what is f′(x)?
- 12x³ − 4x + 5 (Correct answer)
- 12x³ − 2x + 5
- 12x⁴ − 4x² + 5
- 3x³ − 4x + 5
Correct answer: 12x³ − 4x + 5
Applying the power rule term by term: d/dx(3x⁴) = 12x³, d/dx(−2x²) = −4x, d/dx(5x) = 5, and the constant −1 differentiates to 0. So f′(x) = 12x³ − 4x + 5.
Question 2: The derivative f′(a) of a function f at a point x = a represents:
- The average rate of change of f over the interval [0, a]
- The instantaneous rate of change of f at x = a (Correct answer)
- The total accumulated change in f from 0 to a
- The area under the curve f(x) up to x = a
Correct answer: The instantaneous rate of change of f at x = a
The derivative at a specific point gives the instantaneous rate of change — the slope of the tangent line to f at that exact point. The average rate of change uses an interval, and area under the curve is integration, not differentiation.
Question 3: Find the x-values of the stationary points of f(x) = x³ − 3x² − 9x + 5.
- x = 1 or x = −3
- x = 3 or x = −1 (Correct answer)
- x = 3 or x = 1
- x = −3 or x = 1
Correct answer: x = 3 or x = −1
Stationary points occur where f′(x) = 0. f′(x) = 3x² − 6x − 9 = 0 → x² − 2x − 3 = 0 → (x − 3)(x + 1) = 0 → x = 3 or x = −1.
Question 4: A rectangular vegetable garden is enclosed using 40 m of fencing, with one side along an existing wall (no fencing needed there). What dimensions give the maximum garden area?
- Length = 10 m, Width = 20 m
- Length = 20 m, Width = 10 m (Correct answer)
- Length = 15 m, Width = 10 m
- Length = 20 m, Width = 20 m
Correct answer: Length = 20 m, Width = 10 m
Let width = x. Then length = 40 − 2x. Area A = x(40 − 2x) = 40x − 2x². dA/dx = 40 − 4x = 0 gives x = 10 m (width). Length = 40 − 20 = 20 m. Maximum area = 200 m².
Question 5: Find the equation of the tangent to f(x) = x² − 3x at the point where x = 2.
- y = x − 4 (Correct answer)
- y = x + 4
- y = 2x − 4
- y = x − 2
Correct answer: y = x − 4
f(2) = 4 − 6 = −2, so the point is (2, −2). f′(x) = 2x − 3, so f′(2) = 1 (the gradient). Tangent: y − (−2) = 1(x − 2) → y = x − 4.
Question 6: For f(x) = 2x³ − 9x² + 12x − 4, a stationary point occurs at x = 1. What is the nature of this stationary point?
- Local minimum, because f″(1) > 0
- Local maximum, because f″(1) < 0 (Correct answer)
- Inflection point, because f″(1) = 0
- Local minimum, because f′(1) = 0
Correct answer: Local maximum, because f″(1) < 0
f′(x) = 6x² − 18x + 12 and f″(x) = 12x − 18. At x = 1: f″(1) = 12 − 18 = −6 < 0. A negative second derivative means the function is concave down at that point, indicating a local maximum.
NBT Mathematics: Differential Calculus
If f(x) = 3x⁴ − 2x² + 5x − 1, what is f′(x)?