Question

If f(x)=x^3-x, what is the average of change of f(x) over the interval [1,5]

Answer 1

Given:

The function is:

`f(x)=x^3-x`

To find:

The average of change of f(x) over the interval [1,5].

Solution:

We have,

`f(x)=x^3-x`

At x=1, we have

`f(1)=(1)^3-(1)`

`f(1)=1-1`

`f(1)=0`

At x=5, we have

`f(5)=(5)^3-(5)`

`f(5)=125-5`

`f(5)=120`

The average of change of f(x) over the interval [a,b] is

`m=(f(b)-f(a))/(b-a)`

Now, the average of change of f(x) over the interval [1,5] is

`m=(f(5)-f(1))/(5-1)`

`m=(120-0)/(4)`

`m=(120)/(4)`

`m=30`

Therefore, the average of change of f(x) over the interval [1,5] is 30.