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

$m=\dfrac{f(b)-f(a)}{b-a}$

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

$m=\dfrac{f(5)-f(1)}{5-1}$

$m=\dfrac{120-0}{4}$

$m=\dfrac{120}{4}$

$m=30$

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