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Math Help - Homework questions

  1. #1
    Junior Member Sammyj's Avatar
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    Homework questions

    Any help is greatly appreciated

    Use the 2nd derivative test when answering these questions. The critical points of the function f(x) are x = -5, x = 0 and x = 3. The first derivative f'(x) is not known. But the second derivative is f"(x)= 3x^2 + 4x -15

    a) Find all the intervals where the function f(x) is increasing and all the intervals where the function f(x) is decreasing. Explain your reasoning.

    x =-5 3(-5)^2 + 4(-5) -15 = -110 Decreasing (-oo, -5)?
    x = 0 3(0)^2 + 4(0) - 15 = -15 Decreaing (-5, 0)?
    x = 3 3(3)^2 + 4(3) - 15 = 24 Increasing (0, 3)?

    b) Find all the intervals where the function f(x) is concave up and all the intervals where the function f(x) is concave down. Explain your reasoning.

    x =-4 3(-4)^2 + 4(-4) -15 = -79 Concave Down (-oo, -3)
    x = 0 3(0)^2 + 4(0) - 15 = -15 Concave Down (-3, 1.6)
    x = 2 3(2)^2 + 4(2) - 15 = 5 Concave Up (1.6, oo)

    c)Identify the points where all the relative minima occur and all the points where the relative maxima occur. Explain your reasoning.

    x= -3 is a relative min
    x = 1.6 is a relative max??
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Sammyj View Post
    Any help is greatly appreciated

    Use the 2nd derivative test when answering these questions. The critical points of the function f(x) are x = -5, x = 0 and x = 3. The first derivative f'(x) is not known. But the second derivative is f"(x)= 3x^2 + 4x -15

    a) Find all the intervals where the function f(x) is increasing and all the intervals where the function f(x) is decreasing. Explain your reasoning.

    x =-5 3(-5)^2 + 4(-5) -15 = -110 Decreasing (-oo, -5)?
    x = 0 3(0)^2 + 4(0) - 15 = -15 Decreaing (-5, 0)?
    x = 3 3(3)^2 + 4(3) - 15 = 24 Increasing (0, 3)?

    b) Find all the intervals where the function f(x) is concave up and all the intervals where the function f(x) is concave down. Explain your reasoning.

    x =-4 3(-4)^2 + 4(-4) -15 = -79 Concave Down (-oo, -3)
    x = 0 3(0)^2 + 4(0) - 15 = -15 Concave Down (-3, 1.6)
    x = 2 3(2)^2 + 4(2) - 15 = 5 Concave Up (1.6, oo)

    c)Identify the points where all the relative minima occur and all the points where the relative maxima occur. Explain your reasoning.

    x= -3 is a relative min
    x = 1.6 is a relative max??
    Show us some of your working did you use the fact that f(x) is concave up or down where the second derivative is positive and negative respectively?
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  3. #3
    Junior Member Sammyj's Avatar
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    I'm not sure what you're asking. If you want me to draw graphs I do not have that capability online.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Sammyj View Post
    I'm not sure what you're asking. If you want me to draw graphs I do not have that capability online.
    Why not find the first derivative and make this five times easier? Consider that f'(x)=\int{f''(x)}dx=x^3+2x^2-15x+C Now to find C...just use what values make the first derivative zero...in other words the critical poitns
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  5. #5
    Junior Member Sammyj's Avatar
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    Critical points/ F"(x) = 3x^2 + 4x - 15 / F(x) is concave
    ______________________________________________

    -5 / 30-20-15= -5 NEG / Down x = -5 is a max

    0 / -15 NEG / Down x = 0 is a max

    3 / 18 + 12 - 15 = 15 POS / Up x = 3 is max

    Like that?
    Last edited by Sammyj; October 26th 2008 at 12:16 PM.
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  6. #6
    Junior Member Sammyj's Avatar
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    Are questions a & b correct?
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