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Zenmo

  • one year ago

Find an equation of the tangent line to the curve at the given point (Calculus Derivatives).

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  1. Zenmo
    • one year ago
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    Question #1

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  2. freckles
    • one year ago
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    ok what did you find for y'?

  3. Zenmo
    • one year ago
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    \[\lim_{h \rightarrow 0}\frac{ f(a+h)-f(a) }{ h }\] = \[\lim_{h \rightarrow 0}\frac{ [(3+h)^3-3(3+h)+1-(3^3-3(3)+1)] }{ h }\] . Putting more work on here second

  4. FireKat97
    • one year ago
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    do you have to apply differentiation by first principles? :/

  5. Zenmo
    • one year ago
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    \[(3+h)^3; (3+h)(3+h)=9+3h+3h+h^2=9+6h+h^2 ; (9+6h+h^2)(3+h) = 27+18h+3h^2+9h+6h^2+h^3\]

  6. anonymous
    • one year ago
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    Are you familiar with the power rule of differentiation? \[\frac{ d }{ dx }[x^n]=n*x^{n-1}\]

  7. freckles
    • one year ago
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    you might like pascal's triangle 4th row says 1 3 3 1 \[(3+h)^3=1(3)^3+3(3)^2h+3(3)h^2+h^3\]

  8. Zenmo
    • one year ago
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    Our class isn't on there yet, this is the beginning of calculus for the end of Chapter 2

  9. freckles
    • one year ago
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    \[\lim_{h \rightarrow 0} \frac{3^3+3(3)^2h+3(3)h^2+h^3-3(3+h)+1-(3^3-3(3)+1)}{h}\]

  10. Zenmo
    • one year ago
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    Sec. I might found the answer.

  11. Zenmo
    • one year ago
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    I gotten 6x+1 as the answer.

  12. freckles
    • one year ago
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    ok let's go slow then you do see above the 3^3 's cancel right?

  13. freckles
    • one year ago
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    since 3^3-3^3=0 and also you should see the 3(3) also known as 9's cancel you know since -9+9=0

  14. Zenmo
    • one year ago
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    This is how the book does it. Similar problem.

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  15. freckles
    • one year ago
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    you should also see the 1's cancel since 1-1=0

  16. freckles
    • one year ago
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    \[\lim_{h \rightarrow 0} \frac{3^3+3(3)^2h+3(3)h^2+h^3-3(3+h)+1-(3^3-3(3)+1)}{h} \\ \lim_{h \rightarrow 0} \frac{3(3)^2h+3(3)h^2+h^3-3h}{h}\]

  17. freckles
    • one year ago
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    now divide top and bottom by h

  18. Zenmo
    • one year ago
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    \[\lim_{h \rightarrow 0}(h^2+9h+27)\] Did u get that?

  19. freckles
    • one year ago
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    should wind up with \[\lim_{h \rightarrow 0} (3(3)^2+3(3)h+h^2-3)\]

  20. freckles
    • one year ago
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    something happened to your -3 part above

  21. freckles
    • one year ago
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    should be \[\lim_{h \rightarrow 0} (h^2+9h+27-3)\]

  22. freckles
    • one year ago
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    so your slope on your homework is off which will also make your y-intercept you found off too

  23. freckles
    • one year ago
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    please let me know if I need to redo the algebra for you if you don't understand

  24. freckles
    • one year ago
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    I guess you can call it algebra/arithmetic whatever that stuff is

  25. Zenmo
    • one year ago
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    \[\lim_{h \rightarrow 0}(h^2+9h-24)\] that correct? I'm looking at calculations at the moment.

  26. freckles
    • one year ago
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    well 27-3 is 24 not -24 :p

  27. Zenmo
    • one year ago
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    By using the point (3,19) and using the point-slope form for slope M=24. I get \[y=24x+19\]

  28. freckles
    • one year ago
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    y-19=f'(3)(x-3) y-19=24(x-3) y-19=24x-72 hmmm...-72+19...

  29. Zenmo
    • one year ago
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    \[y=24x+53\]

  30. freckles
    • one year ago
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    I hate to be that guy but -72+19 is -53

  31. Zenmo
    • one year ago
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    Yea so the answer is \[y=24x-53\] ?

  32. freckles
    • one year ago
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    yep

  33. freckles
    • one year ago
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    do you understand how we got the slope to be 24?

  34. Zenmo
    • one year ago
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    Yea, turns out I was messing up on simple arithmetic/algebra breaks for: adding/subtracting/factoring etc.

  35. Zenmo
    • one year ago
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    Yea I understood the core concepts, just messed up with the things from above.

  36. Zenmo
    • one year ago
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    Time for a study break haha

  37. Zenmo
    • one year ago
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    Thanks a lot for sticking around and helping. :)

  38. freckles
    • one year ago
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    we all mess on arithmetic just go pet a cat and then come back you will be already (that is my only study advice; cats for the win!)

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