NBT Mathematics: Differential Calculus Flashcards

If f(x) = 3x⁴ − 2x² + 5x − 1, what is f′(x)?

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.

The derivative f′(a) of a function f at a point x = a represents:

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.

Find the x-values of the stationary points of f(x) = x³ − 3x² − 9x + 5.

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.

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?

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.

Find the equation of the tangent to f(x) = x² − 3x at the point where x = 2.

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.

For f(x) = 2x³ − 9x² + 12x − 4, a stationary point occurs at x = 1. What is the nature of this stationary point?

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.