## average rate of change of f(x) = x^2 between x = 1 and x = 4

Simplificatation, evaluation, linear equations, linear graphs, linear inequalities, basic word problems, etc.

### average rate of change of f(x) = x^2 between x = 1 and x = 4

f(x) = x2

1. find the average rate of change of the function f between x=1 and x=4
does this just mean plug in 1for x and plug in 16 for x? and then subtract and get 15?

2. find the value of c making the average rate of change bewteen x=1 and x=c twice your answer to question 1.
does this mean 2 times 15? not sure if this is what its stating. thanks for help.
santaclaus

Posts: 41
Joined: Tue Mar 03, 2009 12:17 am

santaclaus wrote:f(x) = x2

1. find the average rate of change of the function f between x=1 and x=4

Typically, the average rate of change is the absolute change over the interval, divided by the length of the interval. So yes, you'll start by finding the values of f at x = 1 and x = 4. Then you'll subtract these values. This gives you the absolute change. (So you've already done this part.)

Then divide this absolute change over the interval by the length of the interval, to find the average amount of change in f(x) per one-unit change in x.

santaclaus wrote:2. find the value of c making the average rate of change bewteen x=1 and x=c twice your answer to question 1.

Take whatever value you get for (1) and multiply this by 2. (So you should be starting this exercise with a value of 10.)

Now you have to create and solve an equation. So set up the same "average rate of change" process as before, but this time use "c" for the other endpoint:

. . . . .$\frac{f(c)\, -\, f(1)}{c\, -\, 1}\, =\, 10$

. . . . .$\frac{c^2\, -\, 1}{c\, -\, 1}\, =\, 10$

. . . . .$c^2\, -\, 1\, =\, 10(c\, -\, 1)$

Then solve the quadratic equation for the value(s) of "c".

If you get stuck, please reply showing how far you have gotten. Thank you!

stapel_eliz

Posts: 1804
Joined: Mon Dec 08, 2008 4:22 pm