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Math Help - Finding derivatives.

  1. #1
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    Finding derivatives.

    Hello everyone. Once again, I've run into problems in my homework that I was unable to solve, and would greatly appreciate some help in figuring them out. I typed them up on Microsoft Word and screen captured them from there.







    Thank you in advance to whoever helps out.
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  2. #2
    MHF Contributor Unknown008's Avatar
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    1. Well, I don't know that formula, but this is what I do.

    The gradient at any point on a curve is given by the derivative of the curve.

    \dfrac{dy}{dx} = 18x +4

    Now, the gradient is 76. So, the derivative equals 76.

    76 = 18x + 4

    x = 4

    Now, plug in the value of x in the equation of the original curve.

    2. Convert the upper fraction into one single fraction.

    \dfrac{\frac2x - \frac23}{x- 3} = \dfrac{\frac{6-2x}{3x}}{x-3} = \dfrac{2(3-x)}{3x(x-3)} = -\dfrac{2}{3x}

    Try out part b now.

    3. To get the x coordinates of the critical poinst, set the derivative (f'(x) not f(x) as in your post. I think this was a typo) to 0 and solve for x.

    Then, if it is like the cube function, is there a maximum? A minimum perhaps? Or neither?
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  3. #3
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    Quote Originally Posted by Unknown008 View Post
    1. Well, I don't know that formula, but this is what I do.

    The gradient at any point on a curve is given by the derivative of the curve.

    \dfrac{dy}{dx} = 18x +4

    Now, the gradient is 76. So, the derivative equals 76.

    76 = 18x + 4

    x = 4

    Now, plug in the value of x in the equation of the original curve.
    9(x)^2 + 4(x) -> 9(4)^2 + 4(4) = 9(16) + 16 = 160

    Thanks a ton man. I didn't know that all I had to do was get the derivative of the function, and make it equal to 76 to solve for x. I was WAY off the way I was trying to do it before.

    2. Convert the upper fraction into one single fraction.

    \dfrac{\frac2x - \frac23}{x- 3} = \dfrac{\frac{6-2x}{3x}}{x-3} = \dfrac{2(3-x)}{3x(x-3)} = -\dfrac{2}{3x}

    Try out part b now.
    y - (2/3) = -(2/9)(x-3) -> y = -(2/9)x + 6/9 + 2/3 -> y = -2/9x + 4/3

    3. To get the x coordinates of the critical poinst, set the derivative (f'(x) not f(x) as in your post. I think this was a typo) to 0 and solve for x.

    Then, if it is like the cube function, is there a maximum? A minimum perhaps? Or neither?

    Yeah I just noticed when you pointed it out, I did indeed make a typo. It was supposed to be f'(x).

    And the cube function doesn't have a minimum nor a maximum, so then it would be safe to say that there are no critical points on it.
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  4. #4
    MHF Contributor Unknown008's Avatar
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    Yes, for the first 2.

    For 3, there is a critical point, it's only that it's neither a maximum nor a minium but is called an inflection point.
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