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anonymous

  • 5 years ago

find the difference quotient f(x) = 4 +3x-x^2 f(3+h) - f(3)/h please show work

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  1. anonymous
    • 5 years ago
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    the whole thing is over h

  2. amistre64
    • 5 years ago
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    substitute (3+h) and (3) into your function then divide by h and use algebraic techniques to solve

  3. amistre64
    • 5 years ago
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    you end up with 3 - 2x

  4. amistre64
    • 5 years ago
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    if you simply want someone to go thru the effort of typing it all out for you so that you can copy it to your homework; im sure someone will oblige ;)

  5. anonymous
    • 5 years ago
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    not true, I am taking a class and I just don't understand how to do some of the problems. I am a single mom going back to school and struggling a bit. Thanks for your help.

  6. amistre64
    • 5 years ago
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    then at what part does your understanding begin to fade?

  7. amistre64
    • 5 years ago
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    im a single dad going back to college.... so that really doesnt apply does it?

  8. amistre64
    • 5 years ago
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    suppose we have the function: f(x) = x+2 How do we determine: f(5) ??

  9. anonymous
    • 5 years ago
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    Good for you apparently you are better off then I am. I am just looking for help ok?

  10. amistre64
    • 5 years ago
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    or simply put: f(x=5) ?

  11. anonymous
    • 5 years ago
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    You put 5 in for x

  12. amistre64
    • 5 years ago
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    thats right :) then for these that you have we do that same, but with the x=3 and the x=3+h

  13. amistre64
    • 5 years ago
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    what do those look like then?

  14. anonymous
    • 5 years ago
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    so I will end up with 2 answers?

  15. amistre64
    • 5 years ago
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    youll end up subtracting one function from the other ...yes, 4 +3x-x^2

  16. amistre64
    • 5 years ago
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    4 +3(3+h)-(3+h)^2 - [4 +3(x)-(3)^2] = ?

  17. anonymous
    • 5 years ago
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    ok I understand now

  18. amistre64
    • 5 years ago
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    when all that tedious window dressing is done; you should obtain the result; 3+h-2x for an answer :)

  19. anonymous
    • 5 years ago
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    do i multiply by h then to get rid of the h on the bottom?

  20. amistre64
    • 5 years ago
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    you can simply divide everything by h to get rid of it; since there is no = sign to equate to, we should avoid doing the multiply by 'h'.

  21. amistre64
    • 5 years ago
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    it might be correct, but i dont know how accurate it can be :)

  22. anonymous
    • 5 years ago
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    ok

  23. amistre64
    • 5 years ago
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    4 + 9 + 3h -9 -6h -h^2 - 4 -9 + 81 ------------------------------- is what i get so far, if i didnt mess it up h

  24. amistre64
    • 5 years ago
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    gonna have to work it on paper; i get to many typos this way :)

  25. amistre64
    • 5 years ago
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    see, 3^2 = 9; not 81 :)

  26. anonymous
    • 5 years ago
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    The following expression replaces the x in the function value by (h+3) subtract the result of replacing the x in the function value by 3 and then dividing this result by h. \[\left(4+3 x-x^2\text{/.}x\to h+3\right)-\frac{\left(4+3 x-x^2\text{/.}x\to 3\right)}{h} \]= \[9+3 (3+h)-(3+h)^2-3 x \] Although I could be wrong.

  27. amistre64
    • 5 years ago
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    4 + 9 + 3h -9 -6h -h^2 - 4 -9 + 9 ------------------------------- h -3h -h^2 -3h h^2 -------- = ---- - ---- = -3 -h ... but i gotta recheck myself again :) h h h

  28. amistre64
    • 5 years ago
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    3-2x; as x-> 3 we get 3 -6 = -3; ...its good :)

  29. amistre64
    • 5 years ago
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    as h-> 0 we get -3 as an answer, right?

  30. amistre64
    • 5 years ago
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    now how can i make this clearer? or are we good with it?

  31. amistre64
    • 5 years ago
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    robtob; good effort, wrong interpretation of the expression ;)

  32. anonymous
    • 5 years ago
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    amistre64, Blundered on the first post. I think the following is now correct. At 76 years if age I cannot type with the toes of my left foot and monitor what is going on on the computer screen in real time. \[\left(4+3 x-x^2\text{/.}x\to h+3\right)-\frac{\left(4+3 x-x^2\text{/.}x\to 3\right)}{h}=4-\frac{4}{h}-3 h-h^2 \]

  33. amistre64
    • 5 years ago
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    :)

  34. amistre64
    • 5 years ago
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    \[\frac{[4 + 3(3+h) - (3+h)^2] - [4+3(3)-(3)^2]}{h}\]

  35. amistre64
    • 5 years ago
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    since the original equation is : 4+3x-x^2...it has to derive down to: 3 -2x; and at f(3) that has to equal -3 when h=0

  36. amistre64
    • 5 years ago
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    \[\frac{[4+9-9-4+9-9] -3h - h^2}{h}\]

  37. amistre64
    • 5 years ago
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    \[\frac{-3h}{h} - \frac{h^2}{h}\] equals: -3 - h

  38. amistre64
    • 5 years ago
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    when h = 0; this is -3 :)

  39. anonymous
    • 5 years ago
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    thank you so much for your help, sorry my internet dropped

  40. amistre64
    • 5 years ago
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    'sok; internets a fickle beast :)

  41. anonymous
    • 5 years ago
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    amistre64, googled: difference quotient and looked at the first hit. http://answers.yahoo.com/question/index?qid=20090320184937AAl0VC6 Got the general idea now. Thanks.

  42. anonymous
    • 5 years ago
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    I hope i am not in over my head going back to school and starting with a calc class. This site has been really helpful.

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