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anonymous

  • 5 years ago

find the difference quotient of f; that is, find (f(x+h)-f(x))/h f(x)=x^2+5X-1

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  1. anonymous
    • 5 years ago
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    sub in the (x+h) for the values of x in f(x), and subtract f(x). Expand out the f(x+h) and simplify all of your variables. You should end up with\[(h^2+2xh+5h)/h\], the h factors out and you 2x+5, if you take the limit as h goes to zero, which is actually what you are trying to get at in most general purposes.

  2. anonymous
    • 5 years ago
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    so it would be f(x^2+5x-1 +h)-X^2+5x-1 divide by h

  3. anonymous
    • 5 years ago
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    actually it is \[([(x+h)^2 +5(x+h)-1] - [x^2+5x-1]) / h\]. You see, for every value of x in f(x), we are putting x+h in. It is just like putting a constant in like f(5), only it is f(x+h). Does that make sense?. Then you go ahead and expand everything out.

  4. anonymous
    • 5 years ago
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    im not sure how to expand it =(

  5. anonymous
    • 5 years ago
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    okay, the next step is \[([x^2+2xh+h^2+5x+5h-1]-x^2-5x+1)/h\]. You go ahead and square the x + h, and distribute the 5 to (x+h), and make sure to distribute the negative to the f(x) values.

  6. anonymous
    • 5 years ago
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    Do you have any problems factoring or multiplying variables and constants, like (x+5)(x-2), or anything like that?

  7. anonymous
    • 5 years ago
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    sorta

  8. anonymous
    • 5 years ago
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    wats the step after that cuz i got that part

  9. anonymous
    • 5 years ago
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    (2xh+5h+h^2)/h , that turns into 2x + 5 +h after factoring out the h. .If you were to take the limit as I mentioned, you would sub in 0 for h and get 2x+5. Work on getting the factoring and multiplying down, because it's pretty common where you're at right now and onward. If you have any other questions feel free to ask. : )

  10. anonymous
    • 5 years ago
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    thnk you im a fan =)

  11. anonymous
    • 5 years ago
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    No problem, thanks for the nod

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