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onegirl

  • 2 years 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
    • 2 years 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
    • 2 years ago
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    @Mertsj can u help?

  3. zepdrix
    • 2 years ago
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    sup? c: what part confusing you?

  4. onegirl
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    okay

  7. zepdrix
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    k sounds good.

  10. onegirl
    • 2 years 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
    • 2 years ago
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    yes.

  12. onegirl
    • 2 years ago
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    okay

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

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

  15. zepdrix
    • 2 years 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
    • 2 years ago
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    ohhh okay hold on

  17. onegirl
    • 2 years ago
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    so 3e = 3e(1) + b?

  18. zepdrix
    • 2 years ago
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    yes.

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

  20. zepdrix
    • 2 years 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
    • 2 years ago
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    okay

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

  23. zepdrix
    • 2 years 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
    • 2 years ago
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    ok how would i do that?

  25. zepdrix
    • 2 years 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
    • 2 years ago
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    x=x y=y :d

  27. onegirl
    • 2 years ago
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    ok

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