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Math Help - Graph involving integrals and other things

  1. #1
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    Graph involving integrals and other things

    The graph of a function f consists of a semicircle and two line segments as shwon below. Let g(x)=\int^x_1 f(t)dt.
    a.) Find g(1)
    b.) Find g(3)
    c.) Find g(-1)
    d.) Find all values of x on the open interval (-3,4) at which g has a relative maximum.
    e.) Write an equation for the line tangent to the graph of g at x =-1.
    f.) Find the x-coordinate of each point of inflection of the graph of g on the open-interval (-3,4)
    g.) Find the range of g.

    I know how to do the first 3 problems, but the rest are confusing. Any help is appreciated, thanks. Sorry for the bad paint skills...
    Attached Thumbnails Attached Thumbnails Graph involving integrals and other things-ewwww.png  
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  2. #2
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    Let me start you off:

    (d) As x increases from -3 to 1, there is more and more area under the curve. but we're integrating backwards, so the integral is becoming less ans less negative, so g is increasing from -3 to 1.
    As x increases from 1 to 3, the net area decreases (area below the x-axis is negative when integrating), so g is decreasing from 1 to 3.
    As x increases from 3 to 4, the net area increases, so g is increasing from 3 to 4.
    So g has a relative maximum at x=1.

    (e) To write the equation of a line you need a point (x_0,y_0) and a slope m.
    x_0 = -1, y_0=g(-1), m=g'(-1)=f(-1)
    Last edited by DrSteve; January 10th 2011 at 07:32 PM.
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  3. #3
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    I still don't understand the relative maximum question, and I'm having a hard time finding a line tangent to the graph. Help is appreciated, thanks.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by lancelot854 View Post
    I still don't understand the relative maximum question, and I'm having a hard time finding a line tangent to the graph. Help is appreciated, thanks.
    A relative maximum is a point $$x at which f(x)\ge f(y) for all $$y sufficiently close to $$x.

    Thus the peak of the curved part of the graph is a relative maximum (the right hand point of the interval may be considered a relative maximum under some definitions but such end points are usually excluded from definitions of relative extrema).

    CB
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  5. #5
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    Quote Originally Posted by lancelot854 View Post
    I still don't understand the relative maximum question, and I'm having a hard time finding a line tangent to the graph. Help is appreciated, thanks.
    Let me do (d) a different way. The derivative of g is f.
    g'=f is postive on (-3,1) and (3,4). Thus g is increasing on these intervals.
    g'=f is negatve on (1,3). Thus g is decreasing on this interval.
    So g is increasing, then decreasing, then increasing. So g has a relative max at x=1 and a relative min at x=3.

    To finish (e), the point-slope form of an equation of a line is y-y_0=m(x-x_0).

    m=g'(-1)=f(-1)=3, x_0=-1 and y_0=g(x_0)=-\frac{9\pi}{4}.

    So the equation of the tangent line is y+\frac{9\pi}{4}=3(x+1).
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  6. #6
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    Quote Originally Posted by CaptainBlack View Post
    A relative maximum is a point $$x at which f(x)\ge f(y) for all $$y sufficiently close to $$x.

    Thus the peak of the curved part of the graph is a relative maximum (the right hand point of the interval may be considered a relative maximum under some definitions but such end points are usually excluded from definitions of relative extrema).

    CB
    The book says the answer is at x=1, but gives no explanation.
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  7. #7
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    Fortunately you have been given several explanations here. Do you know that there will be a relative max when the derivative of g changes from positive to negative? Do you know what the derivative of g is?
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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by lancelot854 View Post
    The book says the answer is at x=1, but gives no explanation.
    Are you sure it does not say -1?

    CB
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  9. #9
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    The graph of f is the derivative of g , so a relative maximum occurs where f = g' changes from positive to negative (at x = 1).

    note also that the slope of f = slope of g' = g'', which gives the sign of g''

    here's a better picture of the original problem in its entirety ...just shift it left one unit.
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