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anonymous

  • one year ago

Precalculus help!! 1. Find the derivative of f(x) = 6 divided by x at x = -2. and 2. Find the derivative of f(x) = 4x + 7 at x = 5. Thank you. Need help solving, will medal.

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  1. freckles
    • one year ago
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    I bet you are practicing using the formal definition and not the short cuts right?

  2. anonymous
    • one year ago
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    Yes @freckles

  3. freckles
    • one year ago
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    you could use either one of these: \[f'(a)=\lim_{h \rightarrow 0}\frac{f(a+h)-f(a)}{h} \text{ or } f'(a)=\lim_{x \rightarrow a} \frac{f(x)-f(a)}{x-a}\]

  4. freckles
    • one year ago
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    they are the same definition just written a little different

  5. freckles
    • one year ago
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    which one are you more comfortable with?

  6. anonymous
    • one year ago
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    I'm not sure, I'm still confused with both.

  7. freckles
    • one year ago
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    Ok we could do question one together both ways and you can decide which way you want to use for the second and I will check your work for that one

  8. anonymous
    • one year ago
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    Ok thanks!

  9. freckles
    • one year ago
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    \[a=-2 \\ f(x)=\frac{6}{x} \\ f(a)=f(-2)=\frac{6}{-2}=-3 \\ f(a+h)=f(-2+h)=\frac{6}{-2+h}\] first do you understand how I got f(a+h) and f(a) for the first one?

  10. anonymous
    • one year ago
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    Yes completely.

  11. freckles
    • one year ago
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    Ok I'm going to input them into the formula

  12. freckles
    • one year ago
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    \[f'(a)=f'(-2)=\lim_{h \rightarrow 0}\frac{\frac{6}{-2+h}-(-3)}{h}\]

  13. freckles
    • one year ago
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    \[f'(-2)=\lim_{h \rightarrow 0}\frac{\frac{6}{-2+h}+3}{h}\]

  14. freckles
    • one year ago
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    ok the trick is to get that h on the bottom to cancel somehow so we can actually plug in h=0

  15. freckles
    • one year ago
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    notice we have a compound fraction

  16. freckles
    • one year ago
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    that is a fraction inside a bigger fraction

  17. anonymous
    • one year ago
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    Hold on super quick, writing this down.

  18. anonymous
    • one year ago
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    Ok got it.

  19. freckles
    • one year ago
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    we want to see if we can "simplify" the fraction that means get rid of the compound fraction notice the little fraction has denominator (-2+h) multiply top and bottom by (-2+h) this will make our compound fraction into a "simple" fraction do you want to try multiplying top and bottom by (-2+h)

  20. freckles
    • one year ago
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    and let me know what the result is just doing that one step

  21. anonymous
    • one year ago
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    Yes

  22. anonymous
    • one year ago
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    Do I multiply the top and bottom of just the little fraction?

  23. freckles
    • one year ago
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    multiply top and bottom by (-2+h) of the big fraction

  24. anonymous
    • one year ago
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    Would it be 9/h(-2+h) ?

  25. freckles
    • one year ago
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    that isn't a bad try ... I think the 9 is just a bit off...should be 3 there \[\lim_{h \rightarrow 0} \frac{\frac{6}{-2+h}+3}{h} \cdot \frac{-2+h}{-2+h}=?\] \[\lim_{h \rightarrow 0} \frac{\frac{6(-2+h)}{-2+h}+3(-2+h)}{h(-2+h)} \\ =\lim_{h \rightarrow 0}\frac{6+3(-2+h)}{h(-2+h)}\]

  26. freckles
    • one year ago
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    3(-2+h)=-6+3h

  27. freckles
    • one year ago
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    \[\lim_{h \rightarrow 0} \frac{6-6+3h}{h(-2+h)}\]

  28. anonymous
    • one year ago
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    Oh ok, I see. So it would be 3h/h(-2+h)?

  29. freckles
    • one year ago
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    \[\lim_{h \rightarrow 0}\frac{3h}{h(-2+h)}\] you should recall h/h=1 (we can say this since h is not 0)

  30. freckles
    • one year ago
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    \[\lim_{h \rightarrow 0}\frac{3 \cancel{h}}{\cancel{h}(-2+h)}\]

  31. freckles
    • one year ago
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    guess what ?

  32. anonymous
    • one year ago
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    So then we are left with 3/-2+h

  33. freckles
    • one year ago
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    we are going to have no problems pluggin in 0 for h now because this function 3/(-2+h) actually exists at h=0

  34. freckles
    • one year ago
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    so what do you think f'(-2)=?

  35. anonymous
    • one year ago
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    But how do you know that the function exists at 0?

  36. freckles
    • one year ago
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    because the bottom is not 0 when h is 0

  37. anonymous
    • one year ago
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    Oh ok got it, so would f'(-2) = 3/-2?

  38. freckles
    • one year ago
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    yes

  39. freckles
    • one year ago
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    or -3/2

  40. freckles
    • one year ago
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    doesn't matter where you put the negative at

  41. anonymous
    • one year ago
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    Could you check my other question once I solve it?

  42. freckles
    • one year ago
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    k.. while you are working it I'm going to do the first one again but use the other formula I wrote

  43. anonymous
    • one year ago
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    Ok (:

  44. anonymous
    • one year ago
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    Is the second one f(5) = 4?

  45. freckles
    • one year ago
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    \[f'(a)=\lim_{x \rightarrow a}\frac{f(x)-f(a)}{x-a} \\ a=-2 \\ f(x)=\frac{6}{x} \\ f(a)=f(-2)=\frac{6}{-2}=-3 \\ f'(a)=f'(-2)=\lim_{x \rightarrow -2} \frac{\frac{6}{x}+3}{x+2} \\ \text{ multiply \top and bottom by } x \\ f'(-2)=\lim_{x \rightarrow -2} \frac{6+3(x)}{(x+2)(x)} \\ f'(-2)=\lim_{x \rightarrow a}\frac{3(x+2)}{(x+2)(x)} \\ \text{ notice} \frac{x+2}{x+2}=1 \text{ for when } x \neq -2 \\ f'(-2)=\lim_{x \rightarrow -2} \frac{3}{x} =\frac{3}{-2} \text{ or } \frac{-3}{2} \text{ or } -\frac{3}{2} \text{ or } -1.5\]

  46. freckles
    • one year ago
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    oops one sec let me check

  47. freckles
    • one year ago
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    "2. Find the derivative of f(x) = 4x + 7 at x = 5." f(5)=4(5)+7=20+7=27

  48. anonymous
    • one year ago
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    Oh ok, I'm used to the first version we used though. But the one you just used also makes a lot of sense. (:

  49. anonymous
    • one year ago
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    Yes, and for f(a+h) I got it = to 27+h

  50. freckles
    • one year ago
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    did you mean f'(5)=4?

  51. anonymous
    • one year ago
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    Yes that's what I meant.

  52. freckles
    • one year ago
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    that is right

  53. anonymous
    • one year ago
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    Could you check this one too? The position of an object at time t is given by s(t) = -8 - 9t. Find the instantaneous velocity at t = 1 by finding the derivative.

  54. freckles
    • one year ago
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    yeah this question is basically the same as the last questions you asked just asked it in word problem form

  55. anonymous
    • one year ago
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    Would it be: f'(1) = 9?

  56. freckles
    • one year ago
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    the velocity=derivative of position v(t)=s'(t) They are using to find s'(1)

  57. freckles
    • one year ago
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    close

  58. freckles
    • one year ago
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    not exactly though

  59. freckles
    • one year ago
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    you are missing a sign

  60. freckles
    • one year ago
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    \[v(1)=s'(1)=\lim_{h \rightarrow 0}\frac{s(1+h)-s(1)}{h}\]

  61. anonymous
    • one year ago
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    -9?

  62. freckles
    • one year ago
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    yes

  63. anonymous
    • one year ago
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    My bad, didn't catch the negative sign.

  64. freckles
    • one year ago
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    are you using short cuts? or the long way?

  65. freckles
    • one year ago
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    because you found that really fast

  66. anonymous
    • one year ago
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    The long way but using mental math

  67. freckles
    • one year ago
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    nice

  68. anonymous
    • one year ago
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    I have the steps written down though, the long way. Would this be ok for an answer: First, I used: s(t) = -8 - 9t and t=1 to solve for (a+h) and (a). My 'a' term would be 1 in this case. f(a) = -8 - 9t = -8 - 9(1) = -17 f(a+h) = -8 - 9t = -8 - 9(1+h) = -17 - 9h Then I used the equation: f'(a) = lim[f(a+h) - f(a)/h] = f'(a) = -17 - 9h - (-17)/h = -9h/h = -9 Thus I found the derivative to be: f'(1) = -9

  69. freckles
    • one year ago
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    s(t) = -8 - 9t notice the slope of s(t) is -9 the derivative of s(t) is -9

  70. freckles
    • one year ago
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    but yeah your teacher probably wants you to use that formal definition of derivative thing but you know we remember from algebra that the slope of f(x)=mx+b is m and we learn from calculus that the slope is the derivative

  71. anonymous
    • one year ago
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    Ok, but is the way I showed my work using the formal definition?

  72. freckles
    • one year ago
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    yes it is pretty too

  73. freckles
    • one year ago
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    you did very well

  74. freckles
    • one year ago
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    only thing I would say and it is really not apart of your work "s(t) = -8 - 9t and t=1 to solve for (a+h) and (a). My 'a' term would be 1 in this case. :" I would say f(a+h) and f(a) but I knew what you meant when you said this

  75. freckles
    • one year ago
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    oops s(a+h) and s(a) since our function name is s not f

  76. freckles
    • one year ago
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    you could change all those f letters to s letters actually

  77. anonymous
    • one year ago
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    Ok thanks!

  78. freckles
    • one year ago
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    np

  79. anonymous
    • one year ago
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    Thanks for all your help!!

  80. anonymous
    • one year ago
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    I will be doing another test similar if you could help me later on? @freckles Maybe in an hour or so.

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