Polynomial Functions and Rates of Change

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AP Precalculus › Polynomial Functions and Rates of Change

Questions 1 - 4
1

A population model is $N(t)=0.3t^3-2t^2+4t+10$; the rate is $N'(t)$. What is the instantaneous rate of change at $t=5$?​

$N'(5)=13.0$ thousand per year

$N'(5)=-6.5$ thousand per year

$N'(5)=6.5$ thousand per year

$N'(5)=2.5$ thousand per year

Explanation

This question tests AP Precalculus skills, specifically understanding polynomial functions and rates of change. The population function N(t) = 0.3t³ - 2t² + 4t + 10 models growth over time, where N'(t) represents the instantaneous growth rate. To find N'(5), we first find the derivative: N'(t) = 0.9t² - 4t + 4. Substituting t = 5: N'(5) = 0.9(25) - 4(5) + 4 = 22.5 - 20 + 4 = 6.5 thousand per year. Choice A is correct because it accurately calculates the instantaneous rate of change at t = 5. Choice B (2.5) might result from arithmetic errors or incorrect derivative formulation. To help students: Practice finding derivatives of cubic polynomials with decimal coefficients, double-check arithmetic, and interpret positive rates as population growth.

2

A rocket’s height is $h(t)=-5t^2+30t+2$ (meters), and velocity is $h'(t)$. How does the rate of change change on the interval $1,4$?​

It increases by $30$ m/s over $[1,4]$.

It stays constant at $20$ m/s over $[1,4]$.

It decreases by $30$ m/s over $[1,4]$.

It changes from $-10$ m/s to $20$ m/s over $[1,4]$.

Explanation

This question tests AP Precalculus skills, specifically understanding polynomial functions and rates of change. The height function h(t) = -5t² + 30t + 2 models rocket motion, where h'(t) = -10t + 30 represents velocity. To analyze how the rate changes on [1,4], we calculate: h'(1) = -10(1) + 30 = 20 m/s and h'(4) = -10(4) + 30 = -10 m/s. The velocity changes from 20 m/s to -10 m/s, a decrease of 30 m/s. Choice B is correct because it accurately describes this 30 m/s decrease in velocity over the interval. Choice D incorrectly reverses the initial and final values. To help students: Emphasize evaluating derivatives at interval endpoints, interpret the physical meaning of changing rates, and understand that negative acceleration causes velocity to decrease.

3

Stress on a beam (MPa) at position $x$ meters is $S(x)=2x^3-9x^2+12x+4$; the stress gradient is $S'(x)$. What is the instantaneous rate of change at $x=2$?​

$S'(2)=12$ MPa per meter

$S'(2)=6$ MPa per meter

$S'(2)=0$ MPa per meter

$S'(2)=-6$ MPa per meter

Explanation

This question tests AP Precalculus skills, specifically understanding polynomial functions and rates of change. The stress function S(x) = 2x³ - 9x² + 12x + 4 models beam stress, where S'(x) represents the stress gradient (rate of change with position). To find S'(2), we first find the derivative: S'(x) = 6x² - 18x + 12. Substituting x = 2: S'(2) = 6(4) - 18(2) + 12 = 24 - 36 + 12 = 0 MPa per meter. Choice A is correct because the stress gradient equals zero at x = 2, indicating a critical point where stress is neither increasing nor decreasing. Choice B (6 MPa per meter) might result from partial calculation errors. To help students: Practice finding derivatives systematically, understand that zero derivatives indicate critical points, and verify by factoring S'(x) = 6(x² - 3x + 2) = 6(x-1)(x-2).

4

Profit is $P(t)=t^3-6t^2+9t$ (thousands), and growth rate is $P'(t)$. Identify the polynomial’s critical points based on its rate of change.​

Critical points occur at $t=-1$ and $t=3$.

Critical points occur at $t=1$ and $t=3$.

Critical points occur at $t=0$ and $t=3$.

Critical points occur at $t=1$ and $t=9$.

Explanation

This question tests AP Precalculus skills, specifically understanding polynomial functions and rates of change. The profit function P(t) = t³ - 6t² + 9t has derivative P'(t) = 3t² - 12t + 9, which determines critical points where P'(t) = 0. Factoring: P'(t) = 3(t² - 4t + 3) = 3(t - 1)(t - 3), so P'(t) = 0 when t = 1 or t = 3. Choice A is correct because it identifies both critical points where the growth rate equals zero. Choice B incorrectly includes t = 0, which is not a solution to P'(t) = 0. To help students: Practice factoring quadratic expressions, understand that critical points occur where derivatives equal zero, and verify solutions by substitution back into the derivative.