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Math Help - Critical numbers

  1. #1
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    Critical numbers

    need to find critical numbers of f(x) = x^4(x-1)^3

    so I took the derivative using the product rule

    (x-1)^3*4x^3 + x^4*3(x-1)^2

    worked that out to

    4x^3(x^3-3x^2+3x-1) + 3x^4(x^2-2x+1)

    then I'm kinda lost its obvious one critical point is at 0 but there should be at least 1 more 2 if there is an inflection point. Please help me out I'm really lost on this stuff and these problems are due tomorrow.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by UMStudent View Post
    need to find critical numbers of f(x) = x^4(x-1)^3

    so I took the derivative using the product rule

    (x-1)^3*4x^3 + x^4*3(x-1)^2

    worked that out to

    4x^3(x^3-3x^2+3x-1) + 3x^4(x^2-2x+1)

    then I'm kinda lost its obvious one critical point is at 0 but there should be at least 1 more 2 if there is an inflection point. Please help me out I'm really lost on this stuff and these problems are due tomorrow.
    f(x) = (x^4)(x-1)^3
    => f ' (x) = (4x^3)(x - 1)^3 + 3(x^4)(x - 1)^2
    => f ' (x) = (x^3)(x - 1)^2 * [4(x - 1) + 3x]
    => f ' (x) = (x^3)(x - 1)^2 * [7x - 4]

    for critical numbers we have f ' (x) = 0
    => (x^3)(x - 1)^2 * [7x - 4] = 0
    => (x^3) = 0
    or (x - 1)^2 = 0
    or 7x - 4 = 0

    => x = 0
    or x = 1
    or x = 4/7
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  3. #3
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    I follow your first step there thats what I had for the first derivative but I don't see how you go from the derivative to the simplified function and then from that one to the further simplified function.

    f(x) = (x^4)(x-1)^3
    => f ' (x) = (4x^3)(x - 1)^3 + 3(x^4)(x - 1)^2
    => f ' (x) = (x^3)(x - 1)^2 * [4(x - 1) + 3x]
    => f ' (x) = (x^3)(x - 1)^2 * [7x - 4]
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by UMStudent View Post
    its obvious one critical point is at 0 but there should be at least 1 more 2 if there is an inflection point.
    note that when expanding f(x) you get an x^7 graph, that means by rite you should have 7 zeroes in f(x) and 6 zeroes in f ' (x). how come we only have three? some are repeated, the x^3 yields a triple root, the (x - 1)^2 yields a double root and the 7x - 4 yields one root
    Quote Originally Posted by UMStudent View Post
    I follow your first step there thats what I had for the first derivative but I don't see how you go from the derivative to the simplified function and then from that one to the further simplified function.

    f(x) = (x^4)(x-1)^3
    => f ' (x) = (4x^3)(x - 1)^3 + 3(x^4)(x - 1)^2
    => f ' (x) = (x^3)(x - 1)^2 * [4(x - 1) + 3x]
    => f ' (x) = (x^3)(x - 1)^2 * [7x - 4]
    f(x) = (x^4)(x-1)^3
    => f ' (x) = (4x^3)(x - 1)^3 + 3(x^4)(x - 1)^2
    => f ' (x) = (x^3)(x - 1)^2 * [4(x - 1) + 3x] ............here i factorized, (x^3) and (x - 1) are common to both terms
    => f ' (x) = (x^3)(x - 1)^2 * [7x - 4] .....................here i just simplified what's in the square brackets
    Last edited by ThePerfectHacker; April 10th 2007 at 04:08 PM.
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  5. #5
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    Ah alright thanks I see it now, sometimes its hard to see things when they are written on the computer i think.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by UMStudent View Post
    Ah alright thanks I see it now, sometimes its hard to see things when they are written on the computer i think.
    that's true

    sometimes writing stuff out on paper makes it clearer
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