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Calculus 1 : Rate

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Example Questions

Example Question #2 : Rate Of Change

Find the rate of change of a function f(x)=2x-9 from $x_{1}=3$ to $x_{2}=6$ .

Possible Answers:

$\frac{1}{2}$

Correct answer:

Explanation:

Write the formula for the average rate of change from the interval $[{x_{1},x_{2}}]$ .

$\frac{f(x_2)-f(x_{1})}{x_2-x_1}$

Solve for f(x_2) and f(x_1) .

f(x_2=6)=2(6)-9=3

f(x_1=3)=2(3)-9=-3

Substitute the known values into the formula and solve.

$\frac{f(x_2)-f(x_{1})}{x_2-x_1}=\frac{3-(-3)}{6-3}=\frac{6}{3}=2$

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Example Question #1 : Rate Of Change

Suppose the rate of a square is increasing at a constant rate of meters per second. Find the area's rate of change in terms of the square's perimeter.

Possible Answers:

$\frac{1}{2}P$

$\frac{1}{4}P$

$2\sqrt2P$

Correct answer:

Explanation:

Since the question is asking for the rate of change in terms of the perimeter, write the formula for the perimeter of the square and differentiate it with the respect to time.

P=4s

$\frac{dP}{dt}= 4\frac{ds}{dt}$

The question asks in terms of the perimeter. Isolate the term by dividing four on both sides.

$s=\frac{P}{4}$

Write the given rate in mathematical terms and substitute this value into $\frac{dP}{dt}$ .

$\frac{ds}{dt}=2$

$\frac{dP}{dt}= 4\frac{ds}{dt}=4(2)=8$

Write the area of the square and substitute the side.

$A=s^2=\left(\frac{P}{4}\right)^2=\frac{P^2}{16}$

Since the area is changing with time, take the derivative of the area with respect to time.

$\frac{dA}{dt}=\frac{P}{8} \cdot \frac{dP}{dT}$

Substitute the value of $\frac{dP}{dt}$ .

$\frac{dA}{dt}=\frac{P}{8} \cdot 8=P$

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Example Question #1 : Rate Of Change

Determine the point on the function that is not changing: y=x^2-36x+6

Possible Answers:

(0,6)

(18,6)

(36.6)

(6,36)

(18,-318)

Correct answer:

(18,-318)

Explanation:

In order to determine where the function is not changing, it is necessary to take the derivative and set the slope equal to zero. This will provide information on where the curve is not changing. Once we find the x value that gives the derivative a slope of zero, we can substitute the x-value back into the original function to obtain the point.

y=x^2-36x+6

y'=2x-36

0=2x-36

36=2x

x=18

Substitute this value back to the original equation to solve for .

y=18^2-36(18)+6 = 324-648+6 = -318

The point where the function is not changing is (18,-318) .

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Example Question #3 : Rate Of Change

For the function y=10 , what is the average rate of change from x=1 to x=3 ?

Possible Answers:

Correct answer:

Explanation:

Write the formula for average rate of change.

$\frac{f(x_f)-f(x_i)}{x_f-x_i}$

Determine the values of x_f and x_i .

f(x_f)=f(x_i)=10

x_i=1

x_f=3

Substitute the known values.

$\frac{f(x_f)-f(x_i)}{x_f-x_i}=\frac{10-10}{3-1}=0$

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Example Question #4 : Rate Of Change

Find the rate of change of a function f(x)=5x+9 from x_1=4 to x_2=6 .

Possible Answers:

Correct answer:

Explanation:

We can solve by utilizing the formula for the average rate of change: $\frac{f(x_2)-f(x_1)}{x_2-x_1}.$ Solving for f(x) at our given points:

f(x_1)=f(4)=5(4)+9=20+9=29

f(x_2)=f(6)=5(6)+9=30+9=39

Plugging our values into the average rate of change formula, we get:

$\frac{(39-29)}{(6-4)}=\frac{10}{2}=5$

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Example Question #5 : Rate Of Change

Find the rate of change of a function f(x)=9x+7 from x_1=7 to x_2=8

Possible Answers:

Correct answer:

Explanation:

We can solve by utilizing the formula for the average rate of change: $\frac{f(x_2)-f(x_1)}{x_2-x_1}}$ . Solving for f(x) at our given points:

f(x_1)=f(7)=9(7)+7=63+7=70

f(x_2)=f(8)=9(8)+7=72+7=79

Plugging our values into the average rate of change formula, we get:

$\frac{(79-70)}{(8-7)}=\frac{9}{1}=9$

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Example Question #6 : Rate Of Change

Find the rate of change of a function f(x)=3x+7 from x_1=2 to x_2=3 .

Possible Answers:

Correct answer:

Explanation:

We can solve by utilizing the formula for the average rate of change: $\frac{f(x_2)-f(x_1) }{x_2-x_1}$ . Solving for f(x) at our given points:

f(x_1)=f(2)=3(2)+7=6+7=13

f(x_2)=f(3)=3(3)+7=9+7=16

Plugging our values into the average rate of change formula, we get:

$\frac{(16-13)}{(3-2)}=\frac{3}{1}=3$

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Example Question #3 : Rate Of Change

Find the rate of change of a function f(x)=4x+9 from x_1=1 to x_2=2 .

Possible Answers:

Correct answer:

Explanation:

We can solve by utilizing the formula for the average rate of change:

$\frac{ f(x_2)-f(x_1)}{x_2-x_1}$ .

Solving for f(x) at our given points:

f(x_1)=f(1)=4(1)+9=4+9=13

f(x_2)=f(2)=4(2)+9=8+9=17

Plugging our values into the average rate of change formula, we get:

$\frac{17-13}{2-1}=\frac{4}{1}=4$

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Example Question #2 : Rate Of Change

Find the rate of change of a function f(x)=6x-7 from x_1=5 to x_2=6 .

Possible Answers:

Correct answer:

Explanation:

We can solve by utilizing the formula for the average rate of change:

$\frac{ f(x_2)-f(x_1)}{x_2-x_1}$ .

Solving for f(x) at our given points:

f(x_1)=f(5)=6(5)-7=30-7=23

f(x_2)=f(6)=6(6)-7=36-7=29

Plugging our values into the average rate of change formula, we get:

$\frac{29-23}{6-5}=\frac{6}{1}=6$

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Example Question #11 : How To Find Rate Of Change

Find the rate of change of a function f(x)=7x+11 from x_1=4 to x_2=5 .

Possible Answers:

Correct answer:

Explanation:

We can solve by utilizing the formula for the average rate of change:

$\frac{ f(x_2)-f(x_1)}{x_2-x_1}$ .

Solving for f(x) at our given points:

f(x_1)=f(4)=7(4)+11=28+11=39

f(x_2)=f(5)=7(5)+11=35+11=46

Plugging our values into the average rate of change formula, we get:

$\frac{46-39}{5-4}=\frac{7}{1}=7$

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Calculus 1 : Rate

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #2 : Rate Of Change

Example Question #1 : Rate Of Change

Example Question #1 : Rate Of Change

Example Question #3 : Rate Of Change

Example Question #4 : Rate Of Change

Example Question #5 : Rate Of Change

Example Question #6 : Rate Of Change

Example Question #3 : Rate Of Change

Example Question #2 : Rate Of Change

Example Question #11 : How To Find Rate Of Change

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