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Math Help - Maximum Value

  1. #1
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    Maximum Value

    What is the maximum value of f(x)= (x^4) + (x^3) - ((17/4)x^2) + (1/2)x on the interval [-2,2] ?

    I took the derivative of the function and then I set it to 0, and then I factored to get values for x. I got positive and negative 1.09167 and positive 0.060206. However when I plugged it into the original function the answers I were getting were too small. Can someone shed some light for me. Thanks!
    Last edited by abclarinetuvwxyz; January 4th 2009 at 04:48 PM.
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  2. #2
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    Quote Originally Posted by abclarinetuvwxyz View Post
    What is the maximum value of f(x)= (x^4) + (x^3) - ((17/4)x^2) + (1/2)x on the interval [-2,2] ?

    I took the derivative of the function and then I set it to 0, and then I factored to get values for x. I got 0. positive and negative 1.09167 and positive 0.060206. However when I plugged it into the original function the answers I were getting were too small. Can someone shed some light for me. Thanks!
    Do you mean that you got x = 0 as your solution to your y' = 0? I don't think that is correct though.
    Can I ask what the derivative function you obtained?
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  3. #3
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    No. I meant that I then set what I obtained for the derivative to 0. in order to factor.

    What I got for the derivative was :

    f'(x)= 4(x^3) + 3(x^2) - (17/2)x + 1/2
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  4. #4
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    Quote Originally Posted by abclarinetuvwxyz View Post
    No. I meant that I then set what I obtained for the derivative to 0. in order to factor.

    What I got for the derivative was :

    f'(x)= 4(x^3) + 3(x^2) - (17/2)x + 1/2
    The function looks right. And your solutions of 0.060206, 1.09167, -1.09167 looks right too.
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  5. #5
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    Quote Originally Posted by abclarinetuvwxyz View Post
    What is the maximum value of f(x)= (x^4) + (x^3) - ((17/4)x^2) + (1/2)x on the interval [-2,2] ?

    I took the derivative of the function and then I set it to 0, and then I factored to get values for x. I got 0. positive and negative 1.09167 and positive 0.060206. However when I plugged it into the original function the answers I were getting were too small. Can someone shed some light for me. Thanks!
    There is a minimum turning point at x = -1.901875 and x = 1 .09167. There is a maximum turning point at x = 0.060206.

    I don't where your other solutions for f'(x) = 0 have come from ...?

    You should note that over a finite interval the maximum value of a function occurs at either the maximum turning point or one of the endpoints .....
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    I plugged in 0.060206 and I got a decimal, something around 0.014929
    and when I plugged in 1.09167 and I got -1.79783. For this particular problem I know I am supposed to get a number greater than 1 and less than 30. So both of those values do not make sense. Should I try to plug in -1.09167? Can I even do that, is the fact that it is negative significant to this problem?
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    Quote Originally Posted by abclarinetuvwxyz View Post
    I plugged in 0.060206 and I got a decimal, something around 0.014929
    and when I plugged in 1.09167 and I got -1.79783. For this particular problem I know I am supposed to get a number greater than 1 and less than 30. So both of those values do not make sense. Should I try to plug in -1.09167? Can I even do that, is the fact that it is negative significant to this problem?
    *Ahem*
    Quote Originally Posted by mr fantastic View Post
    There is a minimum turning point at x = -1.901875 and x = 1 .09167. There is a maximum turning point at x = 0.060206.

    I don't where your other solutions for f'(x) = 0 have come from ...?

    You should note that over a finite interval the maximum value of a function occurs at either the maximum turning point or one of the endpoints .....
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    Okay thanks, sorry I was typing up my other reply so I just saw your post. Thanks!
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