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brandonloves

  • 2 years ago

Compute and simplify the difference quotient f(a+h)-f(a) --------- h g(x)=2x^2+14x -13

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  1. brandonloves
    • 2 years ago
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    this was already confusing enough without changing it to g(x)

  2. satellite73
    • 2 years ago
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    replace \(g\) by \(f\) it makes no difference, it is all algabra

  3. brandonloves
    • 2 years ago
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    oh ok. so it's like: 2(a+h)^2+14(a)-13 over h?

  4. brandonloves
    • 2 years ago
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    which comes to...

  5. satellite73
    • 2 years ago
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    no hold on

  6. brandonloves
    • 2 years ago
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    oh

  7. satellite73
    • 2 years ago
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    you have to subtract first

  8. satellite73
    • 2 years ago
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    well, first you have to expand, then you have to subtract

  9. satellite73
    • 2 years ago
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    \(g(a+h)= 2(a+h)^2+14(a+h)-13\)

  10. brandonloves
    • 2 years ago
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    expand like 2a+2h.... and so on?

  11. brandonloves
    • 2 years ago
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    (2a+2h)^2?

  12. brandonloves
    • 2 years ago
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    for the first one

  13. satellite73
    • 2 years ago
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    yeah it is \[2(a^2+2ah+h^2)+14a+14h-13\] \[=2a^2+4ah+2h^2+14a+14h-13\]

  14. satellite73
    • 2 years ago
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    square first multiply by 2 second

  15. brandonloves
    • 2 years ago
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    oh ok, gotcha

  16. satellite73
    • 2 years ago
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    from this you need to subtract \(2a^2+14a-13\) so everything without an \(h\) will be gone

  17. satellite73
    • 2 years ago
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    then you can factor out an \(h\) from the remaining terms in the numerator, and cancel with the \(h\) in the denominator

  18. satellite73
    • 2 years ago
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    you should be left with \[4ah +14h+2h^2\] before you divide after you divide you get \[4a+14\] and next week you will be able to do this problem instantly in your head

  19. brandonloves
    • 2 years ago
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    ok, so I have 2a^2 +4ah + 2h^2 +14a +14h -13 and then I subtract 2a2+14a−13? how do I know to do that?

  20. satellite73
    • 2 years ago
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    subtract

  21. satellite73
    • 2 years ago
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    \(2a^2-2a^2=0\) \(14a-14a=0\) and \(-13-(-13)=0\)

  22. satellite73
    • 2 years ago
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    in other words, every term without an \(h\) is gone

  23. brandonloves
    • 2 years ago
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    oh, now I get it

  24. brandonloves
    • 2 years ago
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    and that leaves me with 4ah + 2h^2 +14h

  25. satellite73
    • 2 years ago
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    yes, and now divide by \(h\)

  26. brandonloves
    • 2 years ago
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    as you said lol

  27. brandonloves
    • 2 years ago
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    ok

  28. brandonloves
    • 2 years ago
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    4a+2h+14

  29. brandonloves
    • 2 years ago
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    which is one of my multiple choice answers...

  30. brandonloves
    • 2 years ago
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    well, I dunno if I'll be able to do that in my head instantly, but if I can work it out, then I'll be ok at least haha. Thanks @satellite73

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