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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.....?

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  1. zepdrix
    • one year ago
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    Oh hmm.. this is the other definition of the derivative ok.. lemme try to remember...

  2. zepdrix
    • one year ago
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    \[\large\rm \lim_{x\to a}\frac{f(x)-f(a)}{x-a}\]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
    • one year ago
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    So umm

  4. zepdrix
    • one year ago
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    \[\large\rm f(x)=2x^2+4x+6\]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
    • one year ago
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    \[\large\rm f(-8)=?\]Do you know how to figure this out?

  6. anonymous
    • one year ago
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    yeah i have the answer, it's -28 but every time i plug the expression in it says it's wrong

  7. zepdrix
    • one year ago
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    \[\large\rm \lim_{x\to a}\frac{\color{royalblue}{f(x)}-\color{orangered}{f(-8)}}{x--8}=\lim_{x\to a}\frac{\color{royalblue}{2x^2+4x+6}-(\color{orangered}{2(-8)^2+4(-8)+6})}{x-(-8)}\]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
    • one year ago
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    i did that .-.

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  9. anonymous
    • one year ago
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    well there's an incorrect 2 placed somewhere in there but i changed it and it is still wrong

  10. zepdrix
    • one year ago
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    You still don't have the brackets around the orange like I mentioned.

  11. anonymous
    • one year ago
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    omg

  12. anonymous
    • one year ago
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    you're right! i thought i had put them in lol

  13. anonymous
    • one year ago
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    thank you! it comes out as correct now!

  14. zepdrix
    • one year ago
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    wow that's a lot of brackets <.< lol there must be a better way to do that

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