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Math Help - Concavity

  1. #1
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    Concavity

    For the function f(x)=(x^(2)+4x+(11/4)e^(x) I need to find where it is concave up.

    So, I found the first derivative and then the second derivative which is,
    f''(x)= e^(x)(x^(2)+8x+(51/4))
    Then, I found the roots of the function using the quadratic formula which is (-8+/-sqrt(13))/(2)

    I found it was concave upwards at the open interval: (((-8+sqrt(13))/2), inf)

    But somehow this is incorrect. Can someone help me?
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  2. #2
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    Quote Originally Posted by ctran View Post
    For the function f(x)=(x^(2)+4x+(11/4)e^(x) I need to find where it is concave up.

    So, I found the first derivative and then the second derivative which is,
    f''(x)= e^(x)(x^(2)+8x+(51/4))
    Then, I found the roots of the function using the quadratic formula which is (-8+/-sqrt(13))/(2)

    I found it was concave upwards at the open interval: (((-8+sqrt(13))/2), inf)

    But somehow this is incorrect. Can someone help me?
    is the function ...

    f(x) = \left(x^2 + 4x + \frac{11}{4}\right)e^x<br />

    or

    f(x) = x^2 + 4x + \frac{11}{4}e^x<br />
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  3. #3
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    Quote Originally Posted by skeeter View Post
    is the function ...

    f(x) = \left(x^2 + 4x + \frac{11}{4}\right)e^x<br />

    or

    f(x) = x^2 + 4x + \frac{11}{4}e^x<br />

    Hello skeeter,
    The function is this one,

    f(x) = \left(x^2 + 4x + \frac{11}{4}\right)e^x<br />
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  4. #4
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    Quote Originally Posted by ctran View Post
    Hello skeeter,
    The function is this one,

    f(x) = \left(x^2 + 4x + \frac{11}{4}\right)e^x<br />
    your 2nd derivative is incorrect ...

    f''(x) = e^x\left(x^2+6x+\frac{27}{4}\right)
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  5. #5
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    Quote Originally Posted by skeeter View Post
    your 2nd derivative is incorrect ...

    f''(x) = e^x\left(x^2+6x+\frac{27}{4}\right)


    No I dont think so, f''(x) = e^x\left(x^2+6x+\frac{27}{4}\right) is actually the first derivative.

    I am certain the second derivative is f''(x) = e^x\left(x^2+8x+\frac{51}{4}\right)
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  6. #6
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    Quote Originally Posted by ctran View Post
    No I dont think so, f''(x) = e^x\left(x^2+6x+\frac{27}{4}\right) is actually the first derivative.

    I am certain the second derivative is f''(x) = e^x\left(x^2+8x+\frac{51}{4}\right)
    you're right ... I copied the wrong one down.

    f''(x) = e^x\left(x^2+8x+\frac{51}{4}\right)

    f''(x) = 0 at x = \frac{-8 \pm \sqrt{13}}{2} \approx -2.2 and -5.8


    for x < -5.8 , f''(x) > 0 ... concave up

    for -5.8 < x < -2.8 , f''(x) < 0 ... concave down

    for x > -2.8 , f''(x) > 0 ... concave up
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