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

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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.

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