# Thread: Im not sure if i understand this problem :(

1. ## Im not sure if i understand this problem :(

Let f(x)=8x^2. Find a value A such that the average rate of change of f(x) from 1 to A equals 96.

2. Hello.

The rate of change from $a$ to $b$ is equal to:

$\frac{f(b)-f(a)}{b-a}$

So in your problem, you'd set 96 equal to that. What are $a$ and $b$ in your problem? For that matter, what are $f(a)$ and $f(b)$?

3. Originally Posted by Nox
Hello.

The rate of change from $a$ to $b$ is equal to:

$\frac{f(b)-f(a)}{b-a}$

So in your problem, you'd set 96 equal to that. What are $a$ and $b$ in your problem? For that matter, what are $f(a)$ and $f(b)$?
Im guessing A equals 1 and B is what I have to find.

4. Well, yes. And you may not know exactly what number A is, but you know that it's called "A," so you can use it in the equation. So what are $f(1)$ and $f(A)$?

5. Originally Posted by Nox
Well, yes. And you may not know exactly what number A is, but you know that it's called "A," so you can use it in the equation. So what are $f(1)$ and $f(A)$?
i think f(1)=64 and f(a)=64A

6. Not quite. What you've done is $f(x) = 8^2(x)$ instead of $f(x) = 8(x)^2$.

In other words, square the value of the input and multiply that by 8. Don't square 8 and multiply it by the input, as you have done above.

7. Originally Posted by Nox
Not quite. What you've done is $f(x) = 8^2(x)$ instead of $f(x) = 8(x)^2$.

In other words, square the value of the input and multiply that by 8. Don't square 8 and multiply it by the input, as you have done above.
so that means f(1)=8? i dont understand

8. Yup! $f(1) = 8$. Keep going!

9. Originally Posted by Nox
Yup! $f(1) = 8$. Keep going!
okay....so f(1)=8
which means f(A)=8A^2?

10. Also true. So put those two things back into the equation for rate of change (Remember! Order is important here.) and see what you get.

11. Originally Posted by Nox
Also true. So put those two things back into the equation for rate of change (Remember! Order is important here.) and see what you get.
f(b)-f(a) / b-a
f(b)-8A^2 / b-1

where does the f(1) come into play? i feel slow

12. It's okay, don't feel bad. But $b$ is A, and $a$ is 1. $f(b) - f(a)$ means f(A) - f(1)... in this case, $8A^2 - 8$. Put that over $b - a$. (Which is, in this problem... ??) You're almost there!

13. Originally Posted by Nox
It's okay, don't feel bad. But $b$ is A, and $a$ is 1. $f(b) - f(a)$ means f(A) - f(1)... in this case, $8A^2 - 8$. Put that over $b - a$. (Which is, in this problem... ??) You're almost there!
8A^2-8/A-1?
And then do I set that to 96?
8A^2-8/A-1=96
And i think you multiply both sides by A-1
8A^2-8=96A-96
Is this right so far?

14. Yes! Good work so far, keep going.

15. wow....im stuck. can someone help me?

btw nox, thank you very very much for your help...i appreciate it

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