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Zenmo

  • one year ago

Small Question (formula related). Finding equation of tangent line at the given point.

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  1. Zenmo
    • one year ago
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    Here are the formulas: \[\lim_{x \rightarrow a}\frac{ f(x)-f(a) }{ x-a }\] (#1)

  2. Zenmo
    • one year ago
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    \[\lim_{h \rightarrow 0}\frac{ f(a+h)-f(a) }{ h } \] (#2)

  3. Zenmo
    • one year ago
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    Two example problems: (A) \[y=x^3-3x+1; Point (3,19) \] (B) \[y=\sqrt{x} ; Point (25,5)\]

  4. Zenmo
    • one year ago
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    For problem (A), it uses formula (#2). For problem (B), it uses formula (#1). How do I know what formulas to use, so I won't use formula #1 on example A and formula #2 on example B?

  5. freckles
    • one year ago
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    you can use either 1 or 2

  6. anonymous
    • one year ago
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    yea..

  7. Zenmo
    • one year ago
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    Ah I see, so either can work.

  8. anonymous
    • one year ago
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    i used to get familiar with equation 2 for all the problems when i started calc

  9. anonymous
    • one year ago
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    equation 2 is derived by equation 1. let h=x-a

  10. anonymous
    • one year ago
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    its just a slightly more neater way

  11. anonymous
    • one year ago
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    I'm sure freckles will give a more clear view

  12. freckles
    • one year ago
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    example: \[\lim_{x \rightarrow 3} \frac{f(x)-f(3)}{x-3} \\ =\lim_{x \rightarrow 3} \frac{(x^3-3x+1)-(3^3-3(3)+1)}{x-3} \\ =\lim_{x \rightarrow 3} \frac{(x^3-3^3)+(-3x+3(3))+(1-1)}{x-3 } \\ =\lim_{x \rightarrow 3} \frac{(x-3)(x^2+3x+3^2)-3(x-3)}{x-3} \\ \text{ divide \top and bottom by } x-3\] \[\lim_{x \rightarrow 3} \frac{(x^2+3x+3^2)-3}{1} =\lim_{x \rightarrow 3}(x^2+3x+3^2-3) \\ \text{ plug in 3} \\ 3^2+3(3)+3^2-3=3(3^2)-3=27-3=24\] I hope it is obvious I used the difference of cubes formula above

  13. freckles
    • one year ago
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    oops and that example was for f(x)=x^3-3x+1 at x=3 :

  14. anonymous
    • one year ago
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    \[\lim_{x \rightarrow a}\frac{ f(x)-f(a) }{ x-a }\] let some variable \[h=x-a\] Substitute 'h' into the first equation (note x=h+a) \[\lim_{h-a \rightarrow a}\frac{ f(a+h)-f(a) }{ h }\] \[\lim_{h \rightarrow 0}\frac{ f(a+h)-f(a) }{ h }\]

  15. anonymous
    • one year ago
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    both are inter-related

  16. anonymous
    • one year ago
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    oops

  17. anonymous
    • one year ago
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    i think i made a mistake

  18. Zenmo
    • one year ago
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    second, I need "checking" on this problem.

  19. anonymous
    • one year ago
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    should be h+a==>a

  20. anonymous
    • one year ago
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    \[\lim_{h+a \rightarrow a}\frac{ f(a+h)-f(a) }{ h }\]

  21. freckles
    • one year ago
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    \[\lim_{h \rightarrow 0} \frac{\sqrt{25+h}-\sqrt{25}}{h} \\ \text{ multiply top and bottom by top's conjugate } \\ \lim_{h \rightarrow 0} \frac{\sqrt{25+h}-\sqrt{25}}{h} \cdot \frac{\sqrt{25+h}+\sqrt{25}}{\sqrt{25+h}+\sqrt{25}} \\ \lim_{h \rightarrow 0} \frac{(25+h)-(25)}{h(\sqrt{25+h}+\sqrt{25})} \\ \lim_{h \rightarrow 0} \frac{h}{h(\sqrt{25+h}+\sqrt{25})} \\ \frac{h}{h}=1 \text{ so } \\ \lim_{h \rightarrow 0} \frac{1}{\sqrt{25+h}+\sqrt{25}}=\frac{1}{\sqrt{25}+\sqrt{25}} =\frac{1}{2 \sqrt{25}} =\frac{1}{2(5)}=\frac{1}{10}\] and this is using the other route for the other function you had there at x=25

  22. freckles
    • one year ago
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    though yeah to prove they are the same definition and you can use a substitution like @chris00 is doing above

  23. Zenmo
    • one year ago
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    If I use formula #2 for example B. \[\lim_{h \rightarrow 0}\frac{ f(25+h)-f(25) }{ h }\] =\[\lim_{h \rightarrow 0}\frac{ \sqrt{25+h}-\sqrt{25} }{ h } \] =\[\frac{ \sqrt{25+h}-5 }{ h }\times \frac{ \sqrt{25+h}+5 }{\sqrt{25+h} +5 }\] =\[\lim_{h \rightarrow 0}\frac{ 1 }{ \sqrt{25+h}+5 }\]

  24. freckles
    • one year ago
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    yes!!!! replace h with 0 now

  25. Zenmo
    • one year ago
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    plug in 0 to get 1/10

  26. freckles
    • one year ago
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    yep yep

  27. Zenmo
    • one year ago
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    but for some reason, it counts that as wrong answer

  28. freckles
    • one year ago
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    what is the question?

  29. freckles
    • one year ago
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    exactly

  30. Zenmo
    • one year ago
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  31. Zenmo
    • one year ago
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    Question #2

  32. freckles
    • one year ago
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    it asked for tangent line

  33. freckles
    • one year ago
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    not slope

  34. Zenmo
    • one year ago
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    ops nvm

  35. freckles
    • one year ago
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    but yes slope is required you just need more info

  36. Zenmo
    • one year ago
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    Im derping agn, I got it now, insert into point-slope form to get tangent equation

  37. Zenmo
    • one year ago
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    ok, im going on study break now, Gonna take a step outside haha

  38. Zenmo
    • one year ago
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    Thanks all! :)

  39. freckles
    • one year ago
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    lol good luck

  40. freckles
    • one year ago
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    and sorry didn't mean to give you the answer on that one I kind of thought they had an example or something with that one definition already given

  41. Zenmo
    • one year ago
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    Oh no problem at all, you been greatly helpful so far.

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