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    Need help on this question please - ordinary level calculus. jamielovesmusic123

    Hi There,

    I really need help on this please:

    Let f(x) = x2(x squared) - 3x2 (3x squared) - 24x

    (i) Find the co-ordinates of the local maximum point and the local minimum point on the curve f(x).

    (ii) Show that the curve of f(x) intersects the x-axis at x = 0.

    Thanks in advance. :)

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      Where was this question taken from? Looks like a misprint. Traditionally you're asked to find the local max and min points of either a Cubic or a Quadratic function. The question you're asking is neither.

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      This question was taken from my textbook (Active Maths 3)- Book 1

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      are you sure its 24x and not just 24?

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      I am taking the assumption that your function f(x) = x^3 - 3x^2 - 24x

      (i) First we find the derived function.

      dy/dx = 3x^2 - 6x - 24

      Then let the derived function equal 0 (because the tangents of the local max and local min points are both 0)

      3x^2 - 6x - 24 = 0

      x^2 - 2x - 8 = 0 (dividing both sides by 3)

      (x - 4)(x + 2) = 0

      x = 4 or x = -2

      When x = 4,

      y = (4)^3 - 3(4)^2 - 24(4) = -80

      One of the local points is (4, -80)

      When x = -2

      y = (-2)^3 - 3(-2)^2 - 24(-2) = 28

      The other local point is (-2, 28)

      To determine which of these are local max and min points, we first have to find the second derivative (i.e. the derivative of the derivative of your original function).

      d^2y/dx^2 = 6x - 6

      Then we input the x values of our local points into the second derivative. If the second derivative is negative, the point is a local max. If the second derivative is positive, the point is a local min.

      At x = 4

      6(4) - 6 = 18

      Since it's positive, the point (4, -80) is the local min.

      At x = -2

      6(-2) - 6 = -18

      Since it's negative, the point (-2, 28) is the local max.

      If you're not thought this (the part where I determined which point is local min and which is local max) in OL, you can simply figure that out by looking at the y-value of your two points. Since (-2, 28) has a larger y-value, it will be higher up the co-ordinate plane. Therefore it's the local max. Since (4, -80) has a smaller y-value, it follows that it's the local min.

      (ii) All you have to is input 0 into the function.

      When x = 0, f(x) = (0)^3 - 3(0)^2 - 24(0) = 0

      Since it passes through the point (0, 0) it intersects the x-axis at x = 0. Note that at any point on the x-axis, y = 0.

      Hope this helps :)

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      The start of (ii) is meant to say "All you have to do is input 0 into the function."

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      @Richie_97 if you think about it, the function can't have a constant term (such as 24) if it's going to pass through the x-axis at x = 0 (part (ii) tells us to show this). The constant term acts as the y intercept, as inputting 0 for x will affect all terms (turning them into 0) except for the constant. Generally, for a function ax^3 + bx^2 + cx + d, it will pass through the point (0,d).

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      Thanks everyone for this, It actually really helped :)

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      wat mathes

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      What math

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