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onegirl

  • one year ago

Find an equation of the tangent line to y = f(x) at x = 1. f(x) 3e^x

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  1. onegirl
    • one year ago
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    @zepdrix hi can u help for a minute i know how to do this since you showed me yesterday how..but i'm stuck on something, so can u help?

  2. onegirl
    • one year ago
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    @Mertsj can u help?

  3. zepdrix
    • one year ago
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    sup? c: what part confusing you?

  4. onegirl
    • one year ago
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    okay so i tried doing this one here is what i got so far i found the derivative of 3e^x and i got d/dx (3e^x) = 3(d/dx (e^x)) = 3e^x so now I filled in the 1 into 3e^x and I got 3e so i want to put it into y=mx + b and i got stuck there

  5. zepdrix
    • one year ago
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    Yah it's a weird looking number, so I can see how it's a little confusing. So you calculated the slope, \(\large m\). \(\large m=3e \qquad \rightarrow \qquad y=(3e)x+b\)

  6. onegirl
    • one year ago
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    okay

  7. zepdrix
    • one year ago
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    To solve for \(\large b\) remember that we needed a coordinate pair that we could plug into this. Where can we get a coordinate pair from? :o

  8. onegirl
    • one year ago
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    hmmm 1 in the original question that would be our x and for y we would put 1 into 3e^x

  9. zepdrix
    • one year ago
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    k sounds good.

  10. onegirl
    • one year ago
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    okay so our coordiantes will be (1, 3e) ? since when you plug in 1 into 3e^1 you get 3e

  11. zepdrix
    • one year ago
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    yes.

  12. onegirl
    • one year ago
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    okay

  13. onegirl
    • one year ago
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    so 1 = (3e)x + b ?

  14. onegirl
    • one year ago
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    wait i meant 1 = 3e(1) + b

  15. zepdrix
    • one year ago
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    \[\huge \left(\color{orangered}{x},\;\color{royalblue}{y}\right) \qquad = \qquad \left(\color{orangered}{1},\;\color{royalblue}{3e}\right)\]Plug those into here,\[\huge \color{royalblue}{y}=(3e)\color{orangered}{x}+b\] You didn't quite get them plugged in correctly. Maybe the colors will help you match them up.

  16. onegirl
    • one year ago
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    ohhh okay hold on

  17. onegirl
    • one year ago
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    so 3e = 3e(1) + b?

  18. zepdrix
    • one year ago
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    yes.

  19. onegirl
    • one year ago
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    okay then i solve for be and i'm done right?

  20. zepdrix
    • one year ago
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    Yes, solve for b. Then rewrite your final equatoin with y, x, m and b. Good job c:

  21. onegirl
    • one year ago
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    okay

  22. onegirl
    • one year ago
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    so i fot 3e = 3e(1) + 0

  23. zepdrix
    • one year ago
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    yes, but you want to write it as an equation involving x and y. We no longer want the coordinate pair plugged in for our final answer.

  24. onegirl
    • one year ago
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    ok how would i do that?

  25. zepdrix
    • one year ago
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    Remember we were trying to come up with an equation for the tangent line. It will be of the form \(\large y=mx+b\). We ONLY want to fill in the missing \(\large m\) and \(\large b\). We had plugged in our coordinate pair to help us find \(\large b\), but now we don't need that coordinate pair.

  26. zepdrix
    • one year ago
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    x=x y=y :d

  27. onegirl
    • one year ago
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    ok

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