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anonymous

  • one year ago

Math Problem: (Don't know how to solve this. Please explain when showing how to solve it.)

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  1. anonymous
    • one year ago
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    Evaluate\[(f(x+h)-f(x))/h\] and simplify if \[f(x) = x^2-2x\]

  2. anonymous
    • one year ago
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    this stuff is differentiation

  3. anonymous
    • one year ago
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    i forgot how to do it this method...but i normally use power rule \[\frac{ d }{ dx }x = x ^{n-1}\]

  4. Nnesha
    • one year ago
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    hmm first you need to find f(x+h) so substitute x for (x+h) in f(x) function \[\huge\rm f(x+h)=\color{reD}{x}^2-2\color{ReD}{x}\]

  5. anonymous
    • one year ago
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    so i need to solve for x using that equation u gave me nnesha?

  6. anonymous
    • one year ago
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    and i dont know that power rule saseal

  7. Nnesha
    • one year ago
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    not for x simplify that that's the equation i gave you you already have the formula

  8. Nnesha
    • one year ago
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    and i don't know that either :(

  9. Nnesha
    • one year ago
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    \[\huge\rm f(x+h)=\color{reD}{(x+h)}^2-2\color{ReD}{(x+h)}\] solve that use foil method for (x+h)^2 and distribute parentheses by -2

  10. anonymous
    • one year ago
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    ok...power rule is like (ill just show for x^2) \[\frac{ d }{ dx }(x^2) = 2x\]

  11. anonymous
    • one year ago
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    How did you get 2x? what was the value of d @saseal?

  12. anonymous
    • one year ago
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    actually they call it dy/dx it means to differentiate

  13. anonymous
    • one year ago
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    So the answer for that problem would be\[(x ^{2} +2xh +h ^{2})-2x-2h\] @Nnesha?

  14. anonymous
    • one year ago
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    so are you suppose to plug in y and x instead of d @saseal?

  15. Nnesha
    • one year ago
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    no not the answer that's just f(x+h) now use that formula \[\huge\rm \frac{ f(x+h)-f(x) }{ h }\] replace f(x+h) by (x^2 +2xh+h^2)-2x-2h and f(x) by the function x^2-2x

  16. Nnesha
    • one year ago
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    \[\huge\rm \frac{ \color{ReD}{f(x+h)}-\color{blue}{f(x)} }{ h }\] \[\huge\rm \frac{ \color{reD}{ x^2 +2xh+h^2-2x-2h } -\color{blue}{(x^2 -2x) }}{ h }\] simplify !

  17. anonymous
    • one year ago
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    \[f'(x)=\frac{((x+h)^2-2x+h)- (x^2-2x)}{ h}\]

  18. anonymous
    • one year ago
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    (2xh+h^2-2h)/h

  19. anonymous
    • one year ago
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    That's what I got after putting like terms together

  20. anonymous
    • one year ago
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    ill show you the power rule magic after you get everything together

  21. Nnesha
    • one year ago
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    there is a common factor h^2+2xh+h

  22. anonymous
    • one year ago
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    Alright

  23. anonymous
    • one year ago
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    Oh ya. h(h+2x+1)

  24. anonymous
    • one year ago
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    Is that the final answer?

  25. anonymous
    • one year ago
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    no

  26. anonymous
    • one year ago
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    Am I suppose to factor or insert it into another equation?

  27. Nnesha
    • one year ago
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    \[\large\rm \frac{ \color{reD}{ \cancel{x^2} +2xh+h^2\cancel{-2x}-2h } -\color{blue}{\cancel{x^2} +\cancel{2x }}}{ h }\] \[\large\rm \frac{ h^2+2xh+h }{ h}\] \[\large\rm \frac{ h(h+2x+1)}{ h}\]

  28. Nnesha
    • one year ago
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    can you cancel out h?

  29. anonymous
    • one year ago
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    Oh whoops I forgot about the h on the bottom. So the answer would actually be h+2x+1?

  30. Nnesha
    • one year ago
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    yep

  31. anonymous
    • one year ago
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    Alright thanks nnesha and saseal!

  32. anonymous
    • one year ago
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    where did the coefficient of -2h go?

  33. Nnesha
    • one year ago
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    hmm let's see.

  34. anonymous
    • one year ago
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    I think we wrote that step wrong.

  35. anonymous
    • one year ago
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    Alright its 2x-2+h

  36. Nnesha
    • one year ago
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    \[\large\rm \frac{ \color{reD}{ \cancel{x^2} +2xh+h^2\cancel{-2x}-2h } -\color{blue}{\cancel{x^2} +\cancel{2x }}}{ h }\] \[\large\rm \frac{ h^2+2xh\color{ReD}{-2h} }{ h}\] \[\large\rm \frac{ h(h+2x\color{reD}{-2})}{ h}\] i forgot it suppose to be -2h not just h

  37. Nnesha
    • one year ago
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    thanks saseal

  38. Nnesha
    • one year ago
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    \[\large\rm \frac{ \cancel{h}(h+2x\color{reD}{-2})}{\cancel{ h}}\]

  39. anonymous
    • one year ago
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    Ok, as promised the power rule: you take the power of the variable and pull it down and minus 1 to the power like this \[(\frac{ dy }{ dx })2x^2 = (2)2x ^{2-1}\]

  40. anonymous
    • one year ago
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    Okay

  41. anonymous
    • one year ago
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    if its just 2x it becomes \[\frac{ dy }{ dx }(2x ^{1}) = 2x^0=2(1)\]

  42. anonymous
    • one year ago
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    why did it become 2x^0?

  43. anonymous
    • one year ago
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    so for your equation x^2-2x \[\frac{ dy }{ dx }(x^2-2x) = 2x-2\]

  44. anonymous
    • one year ago
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    because x^1 is x, after you differentiate it you need to -1 from its power so it becomes x^0

  45. anonymous
    • one year ago
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    Okay I think I understand what you did. Would we learn this in AP Calculus AB?

  46. anonymous
    • one year ago
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    if you wonder where the h went...its removed by limits as h approaches 0

  47. anonymous
    • one year ago
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    Yea you would

  48. anonymous
    • one year ago
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    it's like the meat of the calculus

  49. anonymous
    • one year ago
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    Oh okay. Did you take AP Calculus?

  50. anonymous
    • one year ago
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    @saseal

  51. anonymous
    • one year ago
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    No, AP is not available in my country, we do A levels instead which is about the same

  52. anonymous
    • one year ago
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    i don't live in us

  53. anonymous
    • one year ago
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    Oh okay sorry about that.

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