# Thread: Average rate of change

1. ## Average rate of change

Here's the problem: A function is given below. Determine the average rate of change of the function between x = -2 and x = -2 + h.

So what I did was g(-2)= -(5/-2)=5/2
g(-2+h)=-(5/-2+h)= (would this be 5/2+h)

I did this problem a couple of different ways ( I changed around what I thought g(-2+h) would equal) and I have a bunch of different answers. I was wondering if someone could give me the correct answer and explain what they did. Thank you.

2. ## Re: Average rate of change

Average rate of change is how much the function changed divided by how much x changed.

$\displaystyle = \frac{g(-2 + h) - g(-2)}{h}$

= $\displaystyle \frac{\frac{-5}{h-2} - 2.5}{h}$

= $\displaystyle \frac{-5}{h(h-2)} - \frac{2.5}{h}$

= $\displaystyle \frac{-5}{h(h-2)} - \frac{2.5(h-2)}{h(h-2)}$

= $\displaystyle \frac{-5 - 2.5h + 5}{h(h-2)}$

= $\displaystyle \frac{-2.5}{h-2} = \frac{5}{4-2h}$

3. ## Re: Average rate of change

Do you factor out the bottom? 2(2-h)

4. ## Re: Average rate of change

The average rate of change is given by the difference quotient:

$\displaystyle \frac{g(x+h)-g(x)}{h}=\frac{\left(-\frac{5}{x+h} \right)-\left(-\frac{5}{x} \right)}{h}=\frac{-\frac{5}{x+h}+\frac{5}{x}}{h}=$

$\displaystyle \frac{5(x+h)-5x}{hx(x+h)}=\frac{5h}{hx(x+h)}=\frac{5}{x(x+h)}$

Now, plug in -2 for x to get:

$\displaystyle \frac{5}{(-2)((-2)+h)}=\frac{5}{2(2-h)}$

5. ## Re: Average rate of change

I made a common denominator and then I removed an h from both the numerator and denominator after simplifying. Is there a step you want me to clarify?

Edit: if you mean the last step, then I expanded the denominator. $\displaystyle 2(2-h) = 4-2h$

6. ## Re: Average rate of change

Thank you for your help. This is one of the answers I came up with, so I'm glad to know I was doing something right.