Question

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

Answer 1

Answer:

Please check the explanation.

Step-by-step explanation:

Given the function

$f\left(x\right)=\frac{3x+2}{2x-1}$

at x₁ = a,

$f\left(x_1\right)=f\left(a\right)=\frac{3a+2}{2a-1}$

at x₂ = a+h,

$f\left(x_2\right)=f\left(a+h\right)=\frac{3\left(a+h\right)+2}{2\left(a+h\right)-1}$

Using the formula to determine the average rate of change

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

                      $=\:\frac{\frac{3\left(a+h\right)+2}{2\left(a+h\right)-1}-\frac{3a+2}{2a-1}}{a+h-a}\:\:\:\:\:\:\:$

as a+h-a = h, so    

                      $=\frac{\frac{3\left(h+a\right)+2}{2\left(h+a\right)-1}-\frac{3a+2}{2a-1}}{h}$

Thus, the everarge rate of chnage:  $\frac{\frac{3\left(h+a\right)+2}{2\left(h+a\right)-1}-\frac{3a+2}{2a-1}}{h}$

We can further simplify such as:

$=\frac{\frac{3\left(h+a\right)+2}{2\left(h+a\right)-1}-\frac{3a+2}{2a-1}}{h}$

$=\frac{-\frac{7h}{\left(2a-1\right)\left(2\left(h+a\right)-1\right)}}{h}$

$=-\frac{\frac{7h}{\left(2a-1\right)\left(2\left(h+a\right)-1\right)}}{h}$

$=-\frac{7h}{\left(2a-1\right)\left(2\left(h+a\right)-1\right)h}$

Cancel the common factor h

$=-\frac{7}{\left(2a-1\right)\left(2\left(h+a\right)-1\right)}$