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Math Help - Exponential Functions

  1. #1
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    Exponential Functions

    I have had several problems such as the one below, and since e is being raised to a variable, I am getting the same answer for f'(x) and f"(x)...thought I might be doing something incorrectly.

    g(x) = 2 - 3e^x
    g'(x) = -3e^x * (d/dx x)
    g'(x) = -3e^x

    g"(x) = -3e^x (d/dx x)
    g"(x) = -3e^x

    Please either reassure me, or show we where I am mistaken.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by startingover View Post
    I have had several problems such as the one below, and since e is being raised to a variable, I am getting the same answer for f'(x) and f"(x)...thought I might be doing something incorrectly.

    g(x) = 2 - 3e^x
    g'(x) = -3e^x * (d/dx x)
    g'(x) = -3e^x

    g"(x) = -3e^x (d/dx x)
    g"(x) = -3e^x

    Please either reassure me, or show we where I am mistaken.
    you're doing great! though maybe a bit over-zealous...the (d/dx x) is not necessary here. by definition d/dx Ce^x = Ce^x
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    Inflection Pts./Pts of extrema

    How would I go about finding the inflection pts and extrema on a problem such as this? Do you set the problem equal to 0 and solve for e^x?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by startingover View Post
    How would I go about finding the inflection pts and extrema on a problem such as this? Do you set the problem equal to 0 and solve for e^x?
    yes, you would begin by setting the first derivative to zero to find the critical points, then plug those points into the second derivative. if you get a negative result after doing that, it is a local max, if you get a positive, it is a local min, if you get zero, then it is possibly an inflection point, you should test both sides of the critical point to be safe. however, that approach won't work here, as none of the derivatives are ever zero. there are no critical points, so there are no maximums, minimums or inflection points
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