Answer:
Please check the explanation.
Step-by-step explanation:
Given the function
[tex]f\left(x\right)=\frac{3x+2}{2x-1}[/tex]
at x₁ = a,
[tex]f\left(x_1\right)=f\left(a\right)=\frac{3a+2}{2a-1}[/tex]
at x₂ = a+h,
[tex]f\left(x_2\right)=f\left(a+h\right)=\frac{3\left(a+h\right)+2}{2\left(a+h\right)-1}[/tex]
Using the formula to determine the average rate of change
Average rate = [f(x₂) - f(x₁)] / [ x₂ - x₁]
[tex]=\:\frac{\frac{3\left(a+h\right)+2}{2\left(a+h\right)-1}-\frac{3a+2}{2a-1}}{a+h-a}\:\:\:\:\:\:\:[/tex]
as a+h-a = h, so
[tex]=\frac{\frac{3\left(h+a\right)+2}{2\left(h+a\right)-1}-\frac{3a+2}{2a-1}}{h}[/tex]
Thus, the everarge rate of chnage: [tex]\frac{\frac{3\left(h+a\right)+2}{2\left(h+a\right)-1}-\frac{3a+2}{2a-1}}{h}[/tex]
We can further simplify such as:
[tex]=\frac{\frac{3\left(h+a\right)+2}{2\left(h+a\right)-1}-\frac{3a+2}{2a-1}}{h}[/tex]
[tex]=\frac{-\frac{7h}{\left(2a-1\right)\left(2\left(h+a\right)-1\right)}}{h}[/tex]
[tex]=-\frac{\frac{7h}{\left(2a-1\right)\left(2\left(h+a\right)-1\right)}}{h}[/tex]
[tex]=-\frac{7h}{\left(2a-1\right)\left(2\left(h+a\right)-1\right)h}[/tex]
Cancel the common factor h
[tex]=-\frac{7}{\left(2a-1\right)\left(2\left(h+a\right)-1\right)}[/tex]
