## anonymous one year ago Let f(x)=2x2+4x+6. Then according to the definition of derivative f′(−8) is the limit as x tends to -8 of the expression.....?

1. zepdrix

Oh hmm.. this is the other definition of the derivative ok.. lemme try to remember...

2. zepdrix

$\large\rm \lim_{x\to a}\frac{f(x)-f(a)}{x-a}$I believe it looks like that. You have a secant line between the points (x,f(x)) and (a,f(a)) and you let x approach a to get yoor tangent line.

3. zepdrix

So umm

4. zepdrix

$\large\rm f(x)=2x^2+4x+6$The derivative definition evaluated at a=-8 looks like: $\large\rm \lim_{x\to a}\frac{f(x)-f(-8)}{x--8}$So just plug in all the pieces! :)

5. zepdrix

$\large\rm f(-8)=?$Do you know how to figure this out?

6. anonymous

yeah i have the answer, it's -28 but every time i plug the expression in it says it's wrong

7. zepdrix

$\large\rm \lim_{x\to a}\frac{\color{royalblue}{f(x)}-\color{orangered}{f(-8)}}{x--8}=\lim_{x\to a}\frac{\color{royalblue}{2x^2+4x+6}-(\color{orangered}{2(-8)^2+4(-8)+6})}{x-(-8)}$Ya I guess maybe they want you to plug it in completely unsimplified like this? It's important to put brackets around your f(-8) to show that it's being applied to every term over there.

8. anonymous

i did that .-.

9. anonymous

well there's an incorrect 2 placed somewhere in there but i changed it and it is still wrong

10. zepdrix

You still don't have the brackets around the orange like I mentioned.

11. anonymous

omg

12. anonymous

you're right! i thought i had put them in lol

13. anonymous

thank you! it comes out as correct now!

14. zepdrix

wow that's a lot of brackets <.< lol there must be a better way to do that