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Math Help - Intervals

  1. #1
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    Talking Intervals

    need help with few questions:

    1) Find all open intervals on which f(x)=\frac{x^2}{x^2+4} is decreasing.
    this is what i did:
    a. fount the derivative: f'(x)=\frac{8x}{(x^2+4)^2}
    b. set it equal to zero: 8x=0
    c. -{\infty}<x<0 decreasing
    is that all you do? I'm concern for this problem b/c they say to "Find all open intervals".

    2) Consider f(x)=x^{1/2}. Find all value(s), c, in the interval [0,1] such that the slope of the tangent line to the graph of f at c is parallel to the secant line through the point (0, f(0)) and (1, f(1)).
    \frac{1}{2x^{1/2}} i have no idea what to do next...

    3) Explain why f(x)=\frac{1}{x} has a minimum on the interval [1,2] but not on the interval [-1,1].
    graph:

    is it b/c the line is decreasing at interval [1,2]?


    thx in advance
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  2. #2
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    Quote Originally Posted by BryanK View Post
    need help with few questions:

    1) Find all open intervals on which f(x)=\frac{x^2}{x^2+4} is decreasing.
    this is what i did:
    a. fount the derivative: f'(x)=\frac{8x}{(x^2+4)^2}
    b. set it equal to zero: 8x=0
    c. -{\infty}<x<0 decreasing
    is that all you do? I'm concern for this problem b/c they say to "Find all open intervals".
    To find where it is increasing you need to look at the sign of the derivative. Since f'(x) = 8x/(x^2+4)^2 and the denominator is positive the sign only depends on the numerator 8x. Now if x>0 then the numerator is positive and so f'(x)>0 which means f(x) is increasing. Similarly, f(x)<0 for x<0. So it is increasing on (0,\infty) and decreasing (-\infty,0).
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  3. #3
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    Mar 2007
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    um.. u sure that derivative is right?

    i may be horribly mistaken but shouldn’t be:

    x{squared}(x{squared}+4){-1}
    = -x(x{squared}+4){-2}
    :confused:
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