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Math Help - Differentiation

  1. #1
    Member looi76's Avatar
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    Differentiation

    For the following function f(x), find f'(x) and any intervals in which f(x) is increasing:

    x^4 + 4x^3
    f(x) = x^4 + 4x^3
    f'(x) = 4x^3 + 12x^2
    f'(x) = 0
    4x^3+12x^2 = 0
    4x^2(x+3)

    4x^2
    x^2 = \frac{1}{4}
    x = \sqrt{\frac{1}{4}}
    x = 0.5 The answer is wrong. where did I go wrong?

    x+3
    x = -3
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  2. #2
    Bar0n janvdl's Avatar
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    Quote Originally Posted by looi76 View Post
    For the following function f(x), find f'(x) and any intervals in which f(x) is increasing:
    Here you should look at where it has a positive gradient.
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  3. #3
    Bar0n janvdl's Avatar
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    Quote Originally Posted by looi76 View Post
    4x^2 = 0
    x = 0 The answer is wrong. where did I go wrong?
    There you go.
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  4. #4
    Member looi76's Avatar
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    How is it possible? You mean I don't have to substitute x? Every time x would be equal to 0?
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  5. #5
    Bar0n janvdl's Avatar
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    Quote Originally Posted by looi76 View Post
    4x^3+12x^2 = 0
    4x^2(x+3)

    4x^2
    x^2 = \frac{1}{4}
    x = \sqrt{\frac{1}{4}}
    x = 0.5 The answer is wrong. where did I go wrong?

    x+3
    x = -3
    4x^3+12x^2 = 0
    4x^2(x+3) <- Here you threw the "= 0" part away. Don't do that.

    4x^2(x+3) = 0
    4x^2 = 0 and (x+3) = 0
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  6. #6
    Member looi76's Avatar
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    Quote Originally Posted by janvdl View Post
    4x^3+12x^2 = 0
    4x^2(x+3) <- Here you threw the "= 0" part away. Don't do that.

    4x^2(x+3) = 0
    4x^2 = 0 and (x+3) = 0

    for (x+3) = 0, you need to substitute to get x = -3
    but why don't we substitute for 4x^2 = 0 because if we did we would be getting x = 0.5
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  7. #7
    Bar0n janvdl's Avatar
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    Quote Originally Posted by looi76 View Post
    for (x+3) = 0, you need to substitute to get x = -3
    but why don't we substitute for 4x^2 = 0 because if we did we would be getting x = 0.5
    4x^2 = 0

    x^2 = \frac{0}{4}

    x^2 = 0

    x = \sqrt{0}

    x = 0
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  8. #8
    Super Member angel.white's Avatar
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    You have two points: x=-3, and x=0.
    Test the areas around each point. I chose x=-4, x=-1, and x=1

    f\prime(-4) = 4(-4)^3+12(-4)^2 = -64
    f\prime(-1) = 4(-1)^3+12(-1)^2 = 8
    f\prime(1) = 4(1)^3+12(1)^2 = 16

    So we know that before x=-3, the derivative of our function is less than zero, which means that our function is decreasing at this time.
    Between x=-3 and x=0 the derivative becomes positive, so our function is increasing.
    After x= 0 our function is still positive so our function is still increasing.

    This means that our function is increasing at all points after x=-3... EXCEPT that we know the derivative equals zero at the point x=0, so at this point, it is neither increasing nor decreasing.

    So f(x) is increasing on [-3, 0)\cup(0, \infty)

    Quote Originally Posted by looi76 View Post
    for (x+3) = 0, you need to substitute to get x = -3
    but why don't we substitute for 4x^2 = 0 because if we did we would be getting x = 0.5
    No, you would get x=0 as janvdl showed

    You mistakenly set 4x^2 equal to 1, which is where you got 1/4, but you should have kept it equal to zero (because if it equals zero, then it is irrelevant what x+3 equals, since it will be multiplied by zero), and 0/4 = 0, and the square root of 0 = 0 so x=0
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  9. #9
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    Quote Originally Posted by looi76 View Post
    For the following function f(x), find f'(x) and any intervals in which f(x) is increasing:

    x^4 + 4x^3
    f(x) = x^4 + 4x^3
    f'(x) = 4x^3 + 12x^2
    f'(x) = 0
    4x^3+12x^2 = 0
    4x^2(x+3)

    4x^2
    x^2 = \frac{1}{4}
    x = \sqrt{\frac{1}{4}}
    x = 0.5 The answer is wrong. where did I go wrong?

    x+3
    x = -3
    y = f(x) increasing => f'(x) > 0.

    Solving non-linear inequalities is always easiest and safest using a graph. So ....

    Draw a graph of f'(x) = 4x^3 + 12x^2 = 4x^2 (x + 3). You want the values of x for which this graph is ABOVE the x-axis.
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