Instantaneous Rate Of Change

1. The graph shows y = f(x) in purple. Which line has a slope equal to the rate expression shown? $latex=\lim_{h\rightarrow 0}{f(2+h)-f(2)\over h}$ $latex=\lim_{h\rightarrow 0}{f(2+h)-f(2)\over h}$

Line a

Line b

Line c

Line d

Line b has a slope equal to the rate expression shown in the graph. The slope of a line represents the rate of change between two points on the line. In this case, the slope of Line b matches the rate expression shown in the graph, indicating that the function f(x) has a constant rate of change along Line b. Therefore, Line b is the correct answer.

Explanation

Line b has a slope equal to the rate expression shown in the graph. The slope of a line represents the rate of change between two points on the line. In this case, the slope of Line b matches the rate expression shown in the graph, indicating that the function f(x) has a constant rate of change along Line b. Therefore, Line b is the correct answer.

2. Estimate the instantaneous rate using the graph of y = f(x).

0.02

–0.02

7.25

–7.25

The correct answer is -7.25. This is because the graph of y = f(x) has a steep negative slope at the given point, indicating a high rate of change in the negative direction. Therefore, the instantaneous rate at this point is -7.25.

Explanation

The correct answer is -7.25. This is because the graph of y = f(x) has a steep negative slope at the given point, indicating a high rate of change in the negative direction. Therefore, the instantaneous rate at this point is -7.25.

3. Estimate the instantaneous rate of change of H(x) = sin(ln x²) at x = –3

0.1

0.4

0.8

3.5

Undefined

The instantaneous rate of change of a function at a specific point can be found by taking the derivative of the function and evaluating it at that point. In this case, the function H(x) = sin(ln x^2) needs to be differentiated. The derivative of sin(x) is cos(x), and the derivative of ln(x^2) is 2/x. Therefore, the derivative of H(x) is cos(ln x^2) * 2/x. Evaluating this derivative at x = -3 gives cos(ln 9) * 2/-3. Since cos(ln 9) is a constant value, the instantaneous rate of change at x = -3 is 2/(-3) = -2/3. None of the given answer choices match this value, so the correct answer is undefined.

Explanation

The instantaneous rate of change of a function at a specific point can be found by taking the derivative of the function and evaluating it at that point. In this case, the function H(x) = sin(ln x^2) needs to be differentiated. The derivative of sin(x) is cos(x), and the derivative of ln(x^2) is 2/x. Therefore, the derivative of H(x) is cos(ln x^2) * 2/x. Evaluating this derivative at x = -3 gives cos(ln 9) * 2/-3. Since cos(ln 9) is a constant value, the instantaneous rate of change at x = -3 is 2/(-3) = -2/3. None of the given answer choices match this value, so the correct answer is undefined.

4. Acceleration is the rate of change of velocity. If the velocity of a car driving on a straight road is given by v(t), which of the following expressions represent the car's instantaneous expression at t = 5 seconds? (Check all that apply.)

Expression 1:

Expression 2:

Expression 3:

Expression 4:

Expression 5:

Expression 6:

The expressions that represent the car's instantaneous acceleration at t = 5 seconds are Expression 3, Expression 4, and Expression 6.

Explanation

The expressions that represent the car's instantaneous acceleration at t = 5 seconds are Expression 3, Expression 4, and Expression 6.

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5. Which statement(s) correctly contrast average rate of change and instantaneous rate of change?

Average rate of change requires specifying two points on a graph while instantaneous rate of change requires only one.

Average rate of change can be interpreted as a slope while instantaneous rate of change can not.

Average rate of change can be for any function while instantaneous rate of change is only for nonlinear functions.

Average rate of change can be calculated as a difference quotient while instantaneous rate of change is a limit of a difference quotient.

The correct answer states that average rate of change requires specifying two points on a graph while instantaneous rate of change requires only one. This is because average rate of change is calculated by finding the slope between two points on a graph, while instantaneous rate of change is the rate of change at a specific point on a graph. Additionally, the correct answer states that average rate of change can be calculated as a difference quotient, which is the change in the dependent variable divided by the change in the independent variable, while instantaneous rate of change is the limit of a difference quotient as the change in the independent variable approaches zero.

Explanation

The correct answer states that average rate of change requires specifying two points on a graph while instantaneous rate of change requires only one. This is because average rate of change is calculated by finding the slope between two points on a graph, while instantaneous rate of change is the rate of change at a specific point on a graph. Additionally, the correct answer states that average rate of change can be calculated as a difference quotient, which is the change in the dependent variable divided by the change in the independent variable, while instantaneous rate of change is the limit of a difference quotient as the change in the independent variable approaches zero.

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