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Math Help - Differentiation & Stationary Points

  1. #1
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    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
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    Re: Differentiation & Stationary Points

    x-intercepts are where f(x) = 0

    the y-intercept is f(0)

    stationary points are where f'(x) = 0

    also, check your derivative. should be ...

    f'(x) = 3(x^2-1)
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  3. #3
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    Re: Differentiation & Stationary Points

    Hello, smr101!

    You seem to be lacking in much of the basics,
    . . both in Calculus and Algebra.
    Maybe a full solution (with baby steps) will jog your memory.


    Given: . y \:=\:(x-1)^2(x+2)

    (a) Find the coordinates of the points where the curve crosses the x - and y-axes.
    You don't know how to find intercepts?

    Let x = 0:\;y \:=\:(\text{-}1)^2(2) \:=\:2
    . . The y-intercept is (0,2).

    Let y = 0\!:\;0 \:=\:(x-1)^2(x+2) \quad\Rightarrow\quad x \:=\:1,\,\text{-}2
    . . The x-intercepts are (1,0) and (\text{-}2,0).




    (b) Find the stationary points of this curve.

    For the derivative of (a) I got 2x^2 - 3 .This is wrong.
    but I didn't know how to get the points from that. .You don't?

    \text{We have: }\:y \;=\;\overbrace{(x-1)^2}^{f(x)}\overbrace{(x+2)}^{g(x)}

    \text{Product Rule: }\;y' \;=\;f'(x)\!\cdot\!g(x) + f(x)\!\cdot\!g'(x)

    \text{Hence: }\;y' \;=\;2(x-1)\!\cdot\!(x+2) + (x-1)^2\!\cdot\!1 \;=\;2(x^2 +x-2) + x^2-2x+1

    . . . . . . . . =\;2x^2 + 2x - 4 + x^2 - 2x + 1 \;=\;3x^2 - 3

    \text{Then: }\:3x^2 - 3 \:=\:0 \quad\Rightarrow\quad x^2 \:=\:1 \quad\Rightarrow\quad x \:=\:\pm1

    \text{When }x = 1\!:\;\;y = 0
    \text{When }x = \text{-}1\!:\,y = 4

    \text{The stationary points are: }\:(1,0)\text{ and }(\text{-}1,4)
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  4. #4
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    Re: Differentiation & Stationary Points

    Quote Originally Posted by Soroban View Post
    Hello, smr101!

    You seem to be lacking in much of the basics,
    . . both in Calculus and Algebra.
    Maybe a full solution (with baby steps) will jog your memory.



    Let \text{-}1)^2(2) \:=\:2" alt="x = 0:\;y \:=\\text{-}1)^2(2) \:=\:2" />
    . . The y-intercept is (0,2).

    Let x-1)^2(x+2) \quad\Rightarrow\quad x \:=\:1,\,\text{-}2" alt="y = 0\!:\;0 \:=\x-1)^2(x+2) \quad\Rightarrow\quad x \:=\:1,\,\text{-}2" />
    . . The x-intercepts are (1,0) and (\text{-}2,0).





    \text{We have: }\:y \;=\;\overbrace{(x-1)^2}^{f(x)}\overbrace{(x+2)}^{g(x)}

    \text{Product Rule: }\;y' \;=\;f'(x)\!\cdot\!g(x) + f(x)\!\cdot\!g'(x)

    \text{Hence: }\;y' \;=\;2(x-1)\!\cdot\!(x+2) + (x-1)^2\!\cdot\!1 \;=\;2(x^2 +x-2) + x^2-2x+1

    . . . . . . . . =\;2x^2 + 2x - 4 + x^2 - 2x + 1 \;=\;3x^2 - 3

    \text{Then: }\:3x^2 - 3 \:=\:0 \quad\Rightarrow\quad x^2 \:=\:1 \quad\Rightarrow\quad x \:=\:\pm1

    \text{When }x = 1\!:\;\;y = 0
    \text{When }x = \text{-}1\!:\,y = 4

    1,0)\text{ and }(\text{-}1,4)" alt="\text{The stationary points are: }\1,0)\text{ and }(\text{-}1,4)" />
    Thank you!

    It isn't that I don't know how to find y/x intercepts, it's that my derivative answer didn't look right (as you said, it wasn't) which threw me off.
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