Rates of Change
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AP Precalculus › Rates of Change
The average rate of change of the function $$f(x)=x^2$$ is 6. If the interval is $$1, b$$, what is the value of $$b$$?
3
4
5
6
Explanation
The average rate of change is given by $$\frac{f(b) - f(1)}{b - 1} = 6$$. Substituting the function, we get $$\frac{b^2 - 1^2}{b - 1} = 6$$. Factoring the numerator gives $$\frac{(b-1)(b+1)}{b-1} = 6$$. Since $$b \neq 1$$, we can cancel the $$(b-1)$$ terms, leaving $$b+1 = 6$$. Solving for $$b$$ gives $$b=5$$.
Let $$f(x) = -x^3 + 12x + 1$$. On which of the following intervals is $$f(x)$$ increasing at a decreasing rate?
$$(2, \infty)$$
$$(0, 2)$$
$$(-\infty, -2)$$
$$(-2, 0)$$
Explanation
The function is increasing when its rate of change is positive. The rate of change is decreasing when the function is concave down. For a polynomial, this can be analyzed by finding where the function's graph goes up (increasing) and where it is shaped like an upside-down 'U' (concave down). The function increases between its local minimum and maximum, which occur at $$x=-2$$ and $$x=2$$. The point of inflection is at $$x=0$$. The graph is concave down for $$x>0$$. The interval where the function is both increasing and concave down is $$(0, 2)$$.
Let $$f(x) = 2x^2 + 1$$. The rate of change at $$x=1$$ is to be approximated by the average rate of change on the interval $$1, 1.1$$. What is this approximation?
0.42
2.2
4
4.2
Explanation
To find the average rate of change on $$[1, 1.1]$$, we calculate $$f(1.1) = 2(1.1)^2 + 1 = 2(1.21) + 1 = 2.42 + 1 = 3.42$$ and $$f(1) = 2(1)^2 + 1 = 3$$. The average rate of change is $$\frac{3.42 - 3}{1.1 - 1} = \frac{0.42}{0.1} = 4.2$$. This value is an approximation of the instantaneous rate of change at $$x=1$$.
Based on the model, which of the following is the correct interpretation of the average rate of change?
At the beginning of 2015, the fish population was 150 less than it was at the beginning of 2012.
From the beginning of 2012 to the beginning of 2015, the fish population increased on average by 150 fish per year.
From the beginning of 2012 to the beginning of 2015, the total fish population decreased by 150 fish.
From the beginning of 2012 to the beginning of 2015, the fish population decreased on average by 150 fish per year.
Explanation
The average rate of change describes how one quantity changes, on average, with respect to another. The interval $$[2, 5]$$ corresponds to the time from the start of 2012 to the start of 2015. A negative rate of $$-150$$ indicates an average decrease of 150 fish per year over that period.
For which of the following intervals $$a, b$$ is the average rate of change of the function $$p(x) = x^2 - 2x$$ equal to 4?
$$[2, 3]$$
$$[-1, 2]$$
$$[1, 5]$$
$$[0, 4]$$
Explanation
The average rate of change is $$\frac{p(b)-p(a)}{b-a}=4$$. This simplifies to $$\frac{(b^2-2b)-(a^2-2a)}{b-a} = \frac{(b^2-a^2)-2(b-a)}{b-a} = b+a-2$$. We need $$b+a-2 = 4$$, which means $$b+a=6$$. Of the given choices, only the interval $$[1, 5]$$ has endpoints that sum to 6.
Consider the function $$f(x) = \frac{1}{(x-3)^2}$$. How does the rate of change of $$f(x)$$ behave as $$x$$ approaches 3?
The rate of change approaches zero from both the left and the right sides of $$x = 3$$.
The rate of change decreases without bound as $$x$$ approaches 3 from the left and increases without bound as $$x$$ approaches 3 from the right.
The rate of change approaches a large positive value from the left and a large negative value from the right of $$x = 3$$.
The rate of change increases without bound as $$x$$ approaches 3 from the left and decreases without bound as $$x$$ approaches 3 from the right.
Explanation
The graph of $$f(x)$$ has a vertical asymptote at $$x=3$$, with $$f(x) \to \infty$$ from both sides. To the left of the asymptote ($$x<3$$), the function is increasing at an increasingly rapid pace, so its rate of change (slope) increases without bound. To the right of the asymptote ($$x>3$$), the function is decreasing at an increasingly rapid pace, so its rate of change (slope) is negative and decreases without bound.
The polynomial function $$g(x)$$ is decreasing on the interval $$(-\infty, 2)$$ and increasing on the interval $$(2, \infty)$$. Which of the following must be true?
The function $$g(x)$$ has a local minimum at $$x=2$$, and its rate of change switches from negative to positive.
The function $$g(x)$$ has a local maximum at $$x=2$$, and its rate of change switches from negative to positive.
The function $$g(x)$$ has a local maximum at $$x=2$$, and its rate of change switches from positive to negative.
The function $$g(x)$$ has a local minimum at $$x=2$$, and its rate of change switches from positive to negative.
Explanation
A function has a local minimum at a point where it changes from decreasing to increasing. A decreasing function has a negative rate of change, and an increasing function has a positive rate of change. Therefore, at the local minimum at $$x=2$$, the rate of change must switch from negative to positive.
What is the average rate of change of the polynomial function $$f(x) = x^3 - 2x + 1$$ on the interval $$-1, 3$$?
4.5
5
10
20
Explanation
The average rate of change is calculated using the formula $$\frac{f(b) - f(a)}{b - a}$$. First, evaluate the function at the endpoints of the interval: $$f(3) = 3^3 - 2(3) + 1 = 27 - 6 + 1 = 22$$ and $$f(-1) = (-1)^3 - 2(-1) + 1 = -1 + 2 + 1 = 2$$. Then, apply the formula: $$\frac{22 - 2}{3 - (-1)} = \frac{20}{4} = 5$$.
Referring to the context, rate of change means output change per 1-unit input. In a physics test track, a car’s speed is modeled by the polynomial $p(t)= -2t^2+12t$ (m/s) for $0\le t\le 6$. The speed rises from $t=0$ to $t=3$ and then falls from $t=3$ to $t=6$, showing acceleration then deceleration. Over $t=0$ to $t=3$, the average rate of change is $\frac{p(3)-p(0)}{3}$. Over $t=3$ to $t=6$, the average rate of change is $\frac{p(6)-p(3)}{3}$. A sensor delay is modeled by the rational function $r(t)=\frac{10}{t-3}$, which has a vertical asymptote at $t=3$. As $t$ approaches 3, the delay grows without bound in magnitude. Far from $t=3$, the delay approaches 0 seconds. Engineers compare these rates to decide when measurements are least reliable. Based on the passage, at which interval is the car’s average rate of change in speed negative?
From $t=2$ to $t=3$
From $t=0$ to $t=6$
From $t=3$ to $t=6$
From $t=0$ to $t=3$
Explanation
This question tests AP Precalculus understanding of rates of change in polynomial and rational functions. Rates of change measure how a function's output changes relative to its input, crucial for understanding function behavior over intervals. In the given scenario, the car's speed is modeled by p(t) = -2t² + 12t, which increases from t=0 to t=3 (reaching maximum at t=3) and decreases from t=3 to t=6. Choice B is correct because the average rate of change from t=3 to t=6 is [p(6)-p(3)]/3 = [0-18]/3 = -6, which is negative, indicating the car is decelerating. Choice A is incorrect because from t=0 to t=3, the rate is [p(3)-p(0)]/3 = [18-0]/3 = 6, which is positive. Encourage students to calculate average rates systematically using the formula (f(b)-f(a))/(b-a). Watch for: confusing increasing/decreasing function values with positive/negative rates of change.
Referring to the context, rate of change is the average change in speed per second. A car’s speed is modeled by the polynomial $p(t)=-t^2+10t$ (m/s) for $0\le t\le 10$. The speed increases from $t=0$ because the model starts at $p(0)=0$ and rises quickly. It later decreases because the negative quadratic term eventually outweighs the linear term. For example, $p(3)=21$ and $p(7)=21$, indicating the peak occurs between them. A second measurement is modeled by the rational function $r(t)=\frac{50}{t-12}+10$, approaching $10$ as $t$ increases. Near $t=12$, $r(t)$ has a vertical asymptote and is not usable. Based on the passage, what does a negative average rate of change of $p(t)$ on an interval indicate?
The speed becomes undefined at a vertical asymptote on that interval.
The speed equals zero somewhere on that interval.
The speed approaches a horizontal asymptote on that interval.
The speed is decreasing over that time interval.
Explanation
This question tests AP Precalculus understanding of rates of change in polynomial and rational functions. Rates of change measure how a function's output changes relative to its input, crucial for understanding function behavior over intervals. In the given scenario, students must interpret what a negative average rate of change means in the context of speed modeled by p(t) = -t² + 10t. Choice A is correct because a negative average rate of change means the function's output (speed) is decreasing over that time interval - the ending value is less than the starting value. Choice B is incorrect because a negative rate of change doesn't necessarily mean the function equals zero; it only indicates a decrease. Encourage students to connect mathematical concepts to physical interpretations: negative rate of change in speed means deceleration. Practice distinguishing between the sign of a function and the sign of its rate of change.