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Natural exponential Function

  1. Dec 8, 2011 #1
    1. Let f(x)=sin(e^x)

    a. Find 2 values of x satisfying f(x)=0

    b. What is the range of f(x)

    c. Find the value(s) of x that maximize f on [3.8,4] (use calculus)



    2. y=e^x if and only if x=ln y



    3.a. x=-infinity because the limit of e^x as x approaches -infinity is 0. and also, x=infinity because limit e^x as a approaches infinity=infinity

    b.The range of f(x) is (0,infinity)

    c. the values of x that maximize f on 3.81 and 3.95
     
  2. jcsd
  3. Dec 8, 2011 #2

    micromass

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    It might be better in this format:

    1. The problem statement, all variables and given/known data
    Let f(x)=sin(e^x)

    a. Find 2 values of x satisfying f(x)=0

    b. What is the range of f(x)

    c. Find the value(s) of x that maximize f on [3.8,4] (use calculus)

    2. Relevant equations
    y=e^x if and only if x=ln y

    3. The attempt at a solution
    a. x=-infinity because the limit of e^x as x approaches -infinity is 0. and also, x=infinity because limit e^x as a approaches infinity=infinity

    b.The range of f(x) is (0,infinity)

    c. the values of x that maximize f on 3.81 and 3.95[/QUOTE]
     
  4. Dec 9, 2011 #3

    HallsofIvy

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    Thanks, micromass.

    (a) "infiinity" and "-infinity" are NOT "values of x" so I doubt those are acceptable answers. sin(y)= 0 for y any multiple of [itex]\pi[/itex]. You just need to find x such that [itex]e^x= \pi[/itex] and, say, [itex]e^x= 2\pi[/itex].

    (b) Why does the range not include negative numbers? [itex]sin(3\pi/2)= -1[/itex] and there certainly exist x such that [itex]e^x= 3\pi/2[/itex]. And how could [itex]sin(e^x)[/itex] be larger than 1?

    (c) I get more than just two points.
     
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