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Math Help - chain rule and stationary points question

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    chain rule and stationary points question

    I am a bit stuck on a chain rule question and would like to ask for some advice.

    The question starts:

    Q. The graph of y = (x^3 - x^2 +2)^3, is shown (point of inflection at x = -1, max at x=0 and min at unknown x=a point.

    (i) Find the gradient.

    OK so I use chain rule and get dy/dx = 3(x^3 - x^2 +2)^2 (3x^2 - 2x)

    Then next part of question is:

    (ii) Verify, showing your working clearly, that when x = -1, the curve has a point of inflection and when x = 0, the curve has a maximum.

    I tackled this by saying stationary point when dy/dx = 0, then calculated that when x = -1, dy/dx does = 0. Same for x = 0.

    But would I need to do 2nd order differential - ie d^2y/dx^2? Is that required. On above dy/dx is quite hard. Just want to know if there is an easier way.

    Then next part asks:

    (iii) the curve has a minimum when x = a. Find a and verify that this corresponds to a minimum.

    solving this for dy/dx = 0 is quite tricky. How do I solve that? Or is there a way I am missing?

    Angus
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    Re: chain rule and stationary points question

    Quote Originally Posted by angypangy View Post
    I am a bit stuck on a chain rule question and would like to ask for some advice.

    The question starts:

    Q. The graph of y = (x^3 - x^2 +2)^3, is shown (point of inflection at x = -1, max at x=0 and min at unknown x=a point.

    (i) Find the gradient.

    OK so I use chain rule and get dy/dx = 3(x^3 - x^2 +2)^2 (3x^2 - 2x)
    Yep

    Then next part of question is:

    (ii) Verify, showing your working clearly, that when x = -1, the curve has a point of inflection and when x = 0, the curve has a maximum.

    I tackled this by saying stationary point when dy/dx = 0, then calculated that when x = -1, dy/dx does = 0. Same for x = 0.

    But would I need to do 2nd order differential - ie d^2y/dx^2? Is that required. On above dy/dx is quite hard. Just want to know if there is an easier way.
    You'd need to take the second derivative. Thus far you've only proved that -1 is a stationary point, not a point of inflection.

    To take the second derivative use the product rule and the chain rule

    Spoiler:
    u = 3(x^3-x^2+2)^2 \rightarrow u' = 6(x^3-x^2+2)(3x^2-2x)

    v = 3x^2-2x \rightarrow v' = 6x - 2 = 2(3x-1)

    \dfrac{d^2y}{dx^2} = 6(3x^2-2x)^2(x^3-x^2+2) + 6(3x-1)(x^3-x^2+2)^2

    You could simplify but not much point since you're only going to plug and chug


    A point of inflection is when \dfrac{d^2y}{dx^2} = 0 and a maximum is when \dfrac{d^2y}{dx^2} < 0


    Then next part asks:

    (iii) the curve has a minimum when x = a. Find a and verify that this corresponds to a minimum.

    solving this for dy/dx = 0 is quite tricky. How do I solve that? Or is there a way I am missing?

    Angus
    A minimum is when [tex]\dfrac{d^2y}{dx^2} > 0.

    I can see no other way to plug x=a and solve for appropriate values of a. Perhaps someone else will know better?
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