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Math Help - Linear Approx.

  1. #1
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    Linear Approx.

    I know how to do "normal" linear approx...but I don't know how to diff this equation. F(x)= 2^x^2 and approx f(.9) Thank you!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Linnus View Post
    I know how to do "normal" linear approx...but I don't know how to diff this equation. F(x)= 2^x^2 and approx f(.9) Thank you!
    by the chain rule: (or you could notice that 2^{x^2} = e^{x^2 \ln 2})

    \frac d{dx}2^{x^2} = 2x \ln 2 \cdot 2^{x^2} = x \ln 2 \cdot 2^{x^2 + 1}

    the tangent line approximation is given by:

    f(x) \approx f(a) + f'(a)(x - a)

    where x = 0.9 and a = 1


    by "normal" linear approx, i guess you want the normal line. this is just the line perpendicular to the tangent line
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  3. #3
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    Thanks, just disregard the normal thing
    I just want to confirm is this is right..
    f(.9)=2+ln2 (4) (-.1)=2-.4ln2

    Thanks!
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Linnus View Post
    Thanks, just disregard the normal thing
    I just want to confirm is this is right..
    f(.9)=2+ln2 (4) (-.1)=2-.4ln2

    Thanks!
    correct

    the actual answer: 1.7532114...

    our approximation: 1.72274...

    not that bad, but i've seen better, we're only good up to one decimal place here
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