65.3k views
1 vote
Select the correct answer. If f(x)=x^3-x , what is the average rate of change of f(x) over the interval [1, 5]?

User Linog
by
5.9k points

2 Answers

3 votes

Answer:

average rate of change = 30

Explanation:

the average rate of change of f ( x)

over an

interval between 2 points is the slope of the secant

line connecting the 2 points

It is calculated as

f

(

b

)

f

(

a

)

b

a

where a, b is the closed interval

[

a

,

b

]

here

[

a

,

b

]

=

[

1

,

5

]

f

(

b

)

=

f

(

5

)

=

5

3

5

=

120

f

(

a

)

=

f

(

1

)

=

1

1

=

0

120

0

5

1

=

120

4

=

30

User Drmrbrewer
by
5.2k points
5 votes

Answer:

Average rate of change = 30

Explanation:

We have given a function.

f(x) = x³-x

[a,b] = [1,5]

We have to calculate the average rate of change of f(x) over the given interval.

The formula to calculate the average rate of change of function is :

Average rate of change = f(b) - f(a) / (b-a)

f(b) = f(5) = (5)³-5

f(b) = f(5) = 125-5

f(b) = f(5) = 120

f(a) = f(1) = (1)³-1

f(a) = f(1) = 1-1

f(a) = f(1) = 0

Putting values in formula, we have

Average rate of change = 120-0 / 5-1

Average rate of change = 120/4

Average rate of change = 30 which is the answer.

User Oriol Mirosa
by
6.0k points