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The average rate of change of the function f(x)=3sin(x)-1 over the interval [0,…
Related Topics
Wize University Calculus 1 Textbook > Derivatives
Derivatives as a Rate of Change
4 Activities
The average rate of change of the function
f
(
x
)
=
3
s
i
n
(
x
)
−
1
f(x)=3sin(x)-1
f
(
x
)
=
3
s
in
(
x
)
−
1
over the interval
[
0
,
π
6
]
[0, \frac{\pi}{6}]
[
0
,
6
π
]
is
9
9
9
3
3
3
9
π
\frac{9}{\pi}
π
9
0
3
2
π
\frac{3}{2\pi}
2
π
3
I don't know
Check Submission
More Derivatives as a Rate of Change Questions:
Derivatives and Tangent Lines
A certain particle is traveling through space and time. We are able to measure the position of the particle at certain time values. Although we can't measure the particle continuously, we know that the particles position function is differentiable. The table gives values of time for the differentiable function s(t) for
0
≤
t
≤
4
0\le t\le4
0
≤
t
≤
4
. s(t) represents the position of the particle at time t.
Practice: Derivatives as a Rate of Change
f
′
(
x
)
f'(x)
f
′
(
x
)
plot for a differentiable function is depicted as below. Find interval(s) for which
f
(
x
)
f\left(x\right)
f
(
x
)
is increasing.
Average Rate of Change & the Secant Line
Practice: Average Rate of Change & the Secant Line
The population of Vancouver in 2000 was 450,000. In 2020, the population was 740,000. What is the average annual rate of change of the population of Vancouver?
Average Rate of Change & the Secant Line
Practice: Average Rate of Change & the Secant Line
Sherry bought a car in 2010 for
$
60
,
000
\$60,000
$60
,
000
.
50
50
50
years later, she sold the car as a collector car for
$
98
,
000
\$98,000
$98
,
000
. What is the average annual rate of change of the value of the car? (Ignore units)
Average Rate of Change & the Secant Line
Practice: Average Rate of Change & the Secant Line
An investment,
P
P
P
, in $, grows with interest over time
t
t
t
(in years) is given by the table of values below.
t
0
1
2
3
4
P
1000
1305
1528
1774
1936
\begin{array}{|r|c|c|c|c|c|}\hline \textbf{t}&0&1&2&3&4\\\hline \textbf{P}&1000&1305&1528&1774&1936\\\hline \end{array}
t
P
0
1000
1
1305
2
1528
3
1774
4
1936
Average Rate of Change & the Secant Line
Practice: Average Rate of Change & the Secant Line
The amount of bacteria,
A
A
A
, in mg, of a certain species over time
t
t
t
(in minutes) is given by the table of values below.
t
0
1
2
3
4
A
150
183
204
247
283
\begin{array}{|r|c|c|c|c|c|}\hline \textbf{t}&0&1&2&3&4\\\hline \textbf{A}&150&183&204&247&283\\\hline \end{array}
t
A
0
150
1
183
2
204
3
247
4
283
Average Rate of Change & the Secant Line
Practice: Average Rate of Change & the Secant Line
Determine the average rate of the change of the function
g
(
t
)
=
−
3
t
2
t
+
5
g(t)=\dfrac{-3t}{2t+5}
g
(
t
)
=
2
t
+
5
−
3
t
over the interval
0
≤
t
≤
5
0\leq{}t\leq{}5
0
≤
t
≤
5
.
Average Rate of Change & the Secant Line
Practice: Average Rate of Change & the Secant Line
Determine the average rate of the change of the function
g
(
t
)
=
t
2
t
2
−
6
g(t)=\dfrac{t^2}{t^2-6}
g
(
t
)
=
t
2
−
6
t
2
over the interval
0.5
≤
t
≤
2
0.5\leq{}t\leq{}2
0.5
≤
t
≤
2
.
Intervals of Increase and Decrease: Rate of Change
f
′
(
x
)
f'(x)
f
′
(
x
)
plot for a differentiable function is depicted as below. Find interval(s) for which
f
(
x
)
f\left(x\right)
f
(
x
)
is increasing.
The average rate of change of the function
f
(
x
)
=
2
cos
(
x
)
f(x)=2\cos(x)
f
(
x
)
=
2
cos
(
x
)
over the interval
[
π
6
,
π
2
]
[ \frac{\pi}{6},\frac{\pi}{2}]
[
6
π
,
2
π
]
is
Derivatives: Rate of Change
In recent years, exports from Canada to the United States are on the rise. Let
E
(
t
)
E(t)
E
(
t
)
represent the amount of exports to the United States from Canada (in Billions of dollars) at time
t
t
t
. Let
t
t
t
be the number of years after 2000. Interpret the following statements without using the words "derivative" or "instantaneous rate of change".
E
(
2
)
=
5
E(2)=5
E
(
2
)
=
5
E
′
(
3
)
=
1
E'(3)=1
E
′
(
3
)
=
1
Derivatives and Tangent Lines
A certain particle is traveling through space and time. We are able to measure the position of the particle at certain time values. Although we can't measure the particle continuously, we know that the particles position function is differentiable. The table gives values of time for the differentiable function s(t) for
0
≤
t
≤
4
0\le t\le4
0
≤
t
≤
4
. s(t) represents the position of the particle at time t.