Question

Calculate the average rate of change of function on the interval (a,a+h). simplify your expression.Show all steps.

f(x)=3x+2/2x-1

Answer 1

Please check the explanation.

Explanation:

Given the function

`f\left(x\right)=(3x+2)/(2x-1)`

at x₁ = a,

`f\left(x_1\right)=f\left(a\right)=(3a+2)/(2a-1)`

at x₂ = a+h,

`f\left(x_2\right)=f\left(a+h\right)=(3\left(a+h\right)+2)/(2\left(a+h\right)-1)`

Using the formula to determine the average rate of change

Average rate = [f(x₂) - f(x₁)] / [ x₂ - x₁]

`=\:((3\left(a+h\right)+2)/(2\left(a+h\right)-1)-(3a+2)/(2a-1))/(a+h-a)\:\:\:\:\:\:\:`

as a+h-a = h, so

`=((3\left(h+a\right)+2)/(2\left(h+a\right)-1)-(3a+2)/(2a-1))/(h)`

Thus, the everarge rate of chnage:
`((3\left(h+a\right)+2)/(2\left(h+a\right)-1)-(3a+2)/(2a-1))/(h)`

We can further simplify such as:

`=((3\left(h+a\right)+2)/(2\left(h+a\right)-1)-(3a+2)/(2a-1))/(h)`

`=(-(7h)/(\left(2a-1\right)\left(2\left(h+a\right)-1\right)))/(h)`

`=-((7h)/(\left(2a-1\right)\left(2\left(h+a\right)-1\right)))/(h)`

`=-(7h)/(\left(2a-1\right)\left(2\left(h+a\right)-1\right)h)`

Cancel the common factor h

`=-(7)/(\left(2a-1\right)\left(2\left(h+a\right)-1\right))`