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Thread: Differentiation & Stationary Points

  1. #1
    Junior Member
    Sep 2012

    Differentiation & Stationary Points

    Having some trouble with the following two questions, I have an answer but it doesn't make sense...

    A function f is defined by the formula f (x) = (x-1)^2 (x+2)

    (a) Find the coordinates of the points where the curve with the equation y = f(x) crosses the x - and y-axes.

    (b) Find the stationary points of this curve y = f(x).

    For the derivative of (a) I got 2x^2 - 3 but I didn't know how to get the points from that... Help is much appreciated!
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  2. #2
    Aug 2012

    Re: Differentiation & Stationary Points

    The derivative of F(x) is 3X2-5 .....That is your first mistake. unless of course your question writing wasn't correct (i.e. you missed out a + in between the two terms, but as you have written it, your derivative is wrong)
    To find the points where it crosses the y axis you need to start with F(x) and subsitute x=0 to find all the points where it crosses the y axis.
    To find the points crossing the x axis you need to sub in y = 0 and find the values of x that satisfy the equation. I can't tell you how many there are as i am not sure your question writing was correct.

    To find stationary points you need to take the derivative equation and make it equal to zero (i.e. dy/dx = 0 and therefore showing that the gradient of the curve = 0, this is the meaning of a stationary point). You then need to solve the derivative equation for values of x that satisfy, but this time you need to substitute the values you get back into the original F(x) equation to find the value of the y coordinate.

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  3. #3
    MHF Contributor
    Sep 2012
    Washington DC USA

    Re: Differentiation & Stationary Points

    a) The graph of f = { (x,f(x)) in plane | x in R }.
    It intersects the x-axis at points on that graph like (x,0), i.e. where f(x)=0. It intersects the y-axis when x=0, i.e the point (0,f(0)).
    Thus it intersects the x-axis at (1,0) and (-2,0), and interesects the y-axis at (0,2).
    Note it doesn't actually *cross* the x-axis, in the sense of "passing through it" or being transverse to it, at (1,0).

    b) If f(x) = (x-1)^2 (x+2), then f'(x) = d/dx[(x-1)^2] * (x+2) + (x-1)^2 * d/dx[ (x+2) ] = 2(x-1)(x+2) + (x-1)^2 = (x-1) { 2(x+2) + (x-1) } = (x-1)(3x+3) = 3(x-1)(x+1).
    Thus f'(x) = 3(x-1)(x+1) = 3x^2 - 3. Get f'=0 at x=1 and x=-1. Stationary points at x = 1 and x = -1. For the actual point: f(1)=0, f(-1)=4. Thus stationary points are (1,0) and (-1,4).
    Another way to do the derivative - multiply out first: f(x) = (x-1)^2 (x+2) = (x^2-2x+1)(x+2) = x^3 + 0 x^2 - 3x + 2. Thus f'(x) = 3x^2 - 3.
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