Question

f(x)=x^3−9x

What is the average rate of change of f over the interval [1,6]?

Answer 1

Resposta:

Explicação passo-a-passo:

The rate of Change of a function is the derivative of this function:

So the rate of change of f(x) is:

$\dfrac{df(x)}{dx} = \dfrac{dx^3}{dx} - \dfrac{d9x}{dx} = 3x^2 - 9$

In addition, we define the avegare value of a function in an interval by the integral:

$Avarage(f(x)) = \dfrac{1}{b-a}\int\limits^b_a {f(x)} \, dx$

(a,b) is the interval where we are working.

So, We want to know the average value of the rate of change of f(x), then we will integrate f'(x):

$R = \dfrac{1}{6-1}\int\limits^6_1 {f'(x)} \, dx =\dfrac{1}{5}f(x)\limits^6_1 = \dfrac{1}{5}[(6^3 -9.6)-(1^3 - 9.1) ] = \dfrac{1}{5}(216-54-1+9) = \dfrac{170}{5} = 34$

-Note we used the definition of integral like the antiderivate to make the integration.