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Critical points and local extrema Watch

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    The question asks,

    Let g( x)=3-x if x<0, and 3+2x-x^2 if x\ge 0

    Find all the critical points and local extrema.
    How would I go about answering this question?
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    A graph would definitely help. Then any points (e.g. where the two separate graphs meet -- this may or may not be a local extremum, but the graph will make it clear) and stationary points of the quadratic in the region x \ge 0 will be your local extrema.
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    (Original post by nuodai)
    A graph would definitely help. Then any points (e.g. where the two separate graphs meet -- this may or may not be a local extremum, but the graph will make it clear) and stationary points of the quadratic in the region x \ge 0 will be your local extrema.
    Well, basically I've drawn the graph and have identified that the critical point lies within the positive part of the function, that is 3x+2x-x^2.

    So I differentiated the function, which gave me,

    g'( x)=2-2x


    The critical point is when g'( x)=0, so,


    2-2x=0


    2x=2


    x=1 is the critical point.


    Are there any other critical points? Looking at my graph, there doesn't really appear to be any.

    Thanks.
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    (Original post by hollywoodbudgie)
    Well, basically I've drawn the graph and have identified that the critical point lies within the positive part of the function, that is 3x+2x-x^2.

    So I differentiated the function, which gave me,

    g'( x)=2-2x


    The critical point is when g'( x)=0, so,


    2-2x=0


    2x=2


    x=1 is the critical point.


    Are there any other critical points? Looking at my graph, there doesn't really appear to be any.

    Thanks.
    What happens at x=0? [This is the point where, moving from left to right, the graph ceases to be linear and begins to be quadratic.]
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    (Original post by nuodai)
    What happens at x=0? [This is the point where, moving from left to right, the graph ceases to be linear and begins to be quadratic.]
    Isn't it just a local minimum at the point (0,3), with quadratic equation 3+2x-x^2?


    I don't get it?
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    (Original post by hollywoodbudgie)
    Isn't it just a local minimum at the point (0,3), with quadratic equation 3+2x-x^2?


    I don't get it?
    You need to look for points to the left of which the gradient is positive and to the right of which the gradient is negative (or vice versa). This is all a stationary point on a quadratic graph is. Here, we don't have a quadratic graph, we have a graph which is linear for x<0 and quadratic for x \ge 0. The stationary point of the quadratic graph lies at x=1, and so that is one local extremum. There is, however, another extremum, which can't be found by differentiating. When you draw the graph, you'll see that there's a spike that appears at x=0, before which the gradient is negative and after which the gradient is positive.
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    (Original post by nuodai)
    You need to look for points to the left of which the gradient is positive and to the right of which the gradient is negative (or vice versa). This is all a stationary point on a quadratic graph is. Here, we don't have a quadratic graph, we have a graph which is linear for x<0 and quadratic for x \ge 0. The stationary point of the quadratic graph lies at x=1, and so that is one local extremum. There is, however, another extremum, which can't be found by differentiating. When you draw the graph, you'll see that there's a spike that appears at x=0, before which the gradient is negative and after which the gradient is positive.
    I can see the 'downward pointing spike' in my graph, so wouldn't that just mean that x=0 is another critical point?

    So would the answer just be x=1 and x=0?
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    (Original post by hollywoodbudgie)
    I can see the 'downward pointing spike' in my graph, so wouldn't that just mean that x=0 is another critical point?

    So would the answer just be x=1 and x=0?
    Yes (although you might want to give the y-coordinates and say whether it's a local maximum or minimum in each case too). A local extremum is a just bump or spike in the graph.
 
 
 
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