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I think linear approximation? (square root, tangent, e^x)

  1. Feb 2, 2009 #1
    1. The problem statement, all variables and given/known data

    the value of f(x) = (sqrrt e^x +3) at x=0.08 obtained from the tangent to the graph at x=0 is...?



    2. Relevant equations



    3. The attempt at a solution

    i used linear approximation.
    (sqrrt e^o +3) + (1/2(sqrrt3+e^0)(0.08)
    i got an answer but i know its wrong. i got like 1.72 or something.
    did i do it all wrong?
     
  2. jcsd
  3. Feb 2, 2009 #2

    D H

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    Your use of parentheses doesn't make sense. Do you mean

    [tex]f(x) = \sqrt{e^x+3}[/tex]

    [tex]f(x) = \sqrt{e^x} + 3[/tex]

    [tex]f(x) = \left(\sqrt e\right)^{\;x + 3}[/tex]

    [tex]f(x) =\left(\sqrt e\right)^{\;x} + 3 [/tex]

    Or even yet something else?
     
  4. Feb 2, 2009 #3
    yes i meant the first one sorry i dont know how to do that stuff!
     
  5. Feb 2, 2009 #4

    D H

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    You could have written it as f(x)=sqrt(e^x+3) and that would have been fine.
    f(x)=(sqrt e^x+3) was pretty much meaningless.

    What is the derivative of f(x) at x=0?
     
  6. Feb 2, 2009 #5
    would that be 1/2sqrrt(1+3) = 1/4?
    so then i multiply that by 0.08
    and add it to 2.
    ok i got it thanks!
     
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