# Thread: Pre-Calculus Homework Help: Deaing with functions and the average rate of change

1. ## Pre-Calculus Homework Help: Deaing with functions and the average rate of change

I've been stuck on this one problem for a good hour and I just don't understand how to solve it. The directions are as follows: For the following exercises, find the average rate of change of each function on the interval specified for real numbers b or h in simplest form.

11) a(t) = 1/t+4 on [9, 9+h]

I know the answer is -1/13(13+h), though I don't understand why that is. I would appreciate a step-by-step explanation.

2. ## Re: Pre-Calculus Homework Help: Deaing with functions and the average rate of change

Originally Posted by reina
I've been stuck on this one problem for a good hour and I just don't understand how to solve it. The directions are as follows: For the following exercises, find the average rate of change of each function on the interval specified for real numbers b or h in simplest form.

11) a(t) = 1/t+4 on [9, 9+h]

I know the answer is -1/13(13+h), though I don't understand why that is. I would appreciate a step-by-step explanation.
the average rate of change of a function over $[t_1,t_2]$ is given by

$\bar{v} = \dfrac{f(t_2) - f(t_1)}{t_2-t_1}$

in this case

$\bar{v} = \dfrac{\frac{1}{(9+h)+4}-\frac{1}{9+4}}{(9+h)-9} = \dfrac{13-(13+h)}{13(13+h)h} = -\dfrac{1}{13(13+h)}$

3. ## Re: Pre-Calculus Homework Help: Deaing with functions and the average rate of change

Just an addendum to romsek's answer. His "13+ h" is, of course, (9+ h)+ 4= (9+ 4)+ h= 13+ h. Further, to subtract the fractions $\frac{1}{(9+ h)+ 4}- \frac{1}{9+ 4}= \frac{1}{13+ h}- \frac{1}{13}$ you need to get the "common denominator", 13(13+ h) by multiplying numerator and denominator of the $\frac{1}{13+ h}$ by 13 and the numerator and denominator of $\frac{1}{13}$ by 13+ h:
$\frac{13}{13(13+ h)}- \frac{13+ h}{13(13+h)}= \frac{13- 13- h}{13(13+ h)}= \frac{-h}{13(13+ h)}$

Dividing that by (13+ h)- 13= h gives $\frac{-1}{13(13+ h)}$.