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anonymous

  • one year ago

???

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  1. freckles
    • one year ago
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    you can find the tangent line at x=e the only problem is when differentiating f(x)=2ln(x) and f'(e) is probably something you would want to use calculator for but I was told the whole point of these exercises was to approximate without using calculator I guess we could just use the approximate for e as 2.72 when it comes to that anyways the tangent line at x=e will look something like this: y-f(e)=f'(e)(x-e) y=f(e)+f'(e)(x-e) so f(x) is approximately f(e)+f'(e)(x-e)

  2. anonymous
    • one year ago
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    So I would plug in values into that?

  3. freckles
    • one year ago
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    you can use that approximation equation to approximate f(2) f(2) is approximately f(e)+f'(e)(2-e)

  4. freckles
    • one year ago
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    you still need to find f(e) and f'(e) though

  5. anonymous
    • one year ago
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    How would I find those. And would it help if i showed the answer choices?

  6. freckles
    • one year ago
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    there are choices for approximating f(2)?

  7. freckles
    • one year ago
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    f(e) can bound by pluggin in e into f(x)=2ln(x) f'(x) needs to be found before finding f'(e)

  8. anonymous
    • one year ago
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    y = x − 2 y = x − 2 + ln4 y = 2ln(x − 2) y = 1 This is what it shows

  9. freckles
    • one year ago
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    oh they want the linear approximation for f(2)

  10. freckles
    • one year ago
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    they want the tangent line at x=2

  11. freckles
    • one year ago
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    so just find the tangent line at x=2

  12. anonymous
    • one year ago
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    When I derive 2lnx I get 2/x

  13. freckles
    • one year ago
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    right now plug in 2

  14. freckles
    • one year ago
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    \[y-f(2)=f'(2)(x-2) \\ y=f(2)+f'(2)(x-2)\]

  15. anonymous
    • one year ago
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    I got 1 when I plugged it in

  16. freckles
    • one year ago
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    ok so f'(2)=1 you still need to find f(2)

  17. freckles
    • one year ago
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    \[y=f(2)+1(x-2) \\ y=f(2)+(x-2) \\ y=f(2)+x-2 \\ y=x-2+f(2)\]

  18. freckles
    • one year ago
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    f(x)=2ln(x) input 2 to find f(2)

  19. anonymous
    • one year ago
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    1.39

  20. freckles
    • one year ago
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    don't approximate plug in 2

  21. freckles
    • one year ago
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    then just use power rule for log

  22. freckles
    • one year ago
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    \[a \ln(x)=\ln(x^a) \text{ power rule }\]

  23. freckles
    • one year ago
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    \[f(2)=2\ln(2)=...\]

  24. anonymous
    • one year ago
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    When I plugged that into my calculator I got 1.39

  25. freckles
    • one year ago
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    yes that is great but we are not approximating use power rule bring the 2 up as an exponent of what is inside the log

  26. freckles
    • one year ago
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    example \[5\ln(2)=\ln(2^5)=\ln(32)\]

  27. freckles
    • one year ago
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    you try 2ln(2)=what using power rule

  28. anonymous
    • one year ago
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    So I would do ln(2^2)=ln(4)

  29. freckles
    • one year ago
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    yes

  30. freckles
    • one year ago
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    the tangent line at x=a looks like this y-f(a)=f'(a)(x-a) so we has 2 instead of a y-f(2)=f'(2)(x-2) you found f(2) as ln(4) and f'(2) as 1 y-ln(4)=1(x-2) y-ln(4)=x-2 add ln(4) on both sides and you should see clearly see your answer

  31. anonymous
    • one year ago
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    x-2+ln4?

  32. freckles
    • one year ago
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    yes

  33. freckles
    • one year ago
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    y=x-2+ln(4)

  34. anonymous
    • one year ago
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    Thank you so much!!!!

  35. freckles
    • one year ago
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    np

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