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Math Help - Graphing using calculus

  1. #1
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    Graphing using calculus

    Hey all, could someone please check my answers? thanks

    1.) Consider the function k(x) = (x^2+1)arctan(x)

    Useful Information:
    k'(x) = 1+2xarctan(x)
    k''(x) = (2x/x^2+1)+ 2arctan(x)

    __________________________________________________ ______________
    a) Find the domain of k

    my solution: dom(k) = R

    __________________________________________________ ______________

    b) Find x and y intercepts

    my solution x int at (0,0) and y int at (0,0)

    __________________________________________________ ______________

    c) Find the interval on which k is increasing:

    my solution: since k'(x) > 0 on (0,inf), k increases from (0,inf)

    __________________________________________________ ______________

    d) Find the interval on which k is decreasing

    my solution: since k'(x) < 0 on (-inf,0), k decreases from (-inf, 0)

    __________________________________________________ ______________

    e) Does k have any stationary points?

    my solution: k'(x) = 0
    1+2xarctan(x) = 0
    arctan(x) = -(1/2x)
    since the two graphs never intersect, there are no stationary points

    __________________________________________________ _______________

    f) show that k has point of inflection at x = 0

    my solution: k''(0) = 0, therefore point of inflection.

    __________________________________________________ ____________________

    g) Is the point of inflection at x = 0 stationary or non stationary?

    my solution: k'(0) = 1, therefore non stationary point of inflection at x = 0

    __________________________________________________ _____________________

    h) Does k have any other point of inflections?

    my solution: k''(x) = 0
    (2x/1+x^2)+2 arctan(x) = 0
    I got stuck at this point, any guidance would be much appreciated.

    Thanks.
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Graphing using calculus

    This looks fine! There's only one point of inplection (0,0). It's not easy to solve equations like \frac{2x}{1+x^2}+2\arctan(x)=0 because you're dealing with polynomial functions and cyclometric functions at the same time. Offcourse by inspection you notice x=0 is a zero and this is the only zero and thus point of inflection (take a look at the graph).
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  3. #3
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    Re: Graphing using calculus

    parts (c) & (d) appear to be wrong.
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: Graphing using calculus

    Quote Originally Posted by SammyS View Post
    parts (c) & (d) appear to be wrong.
    Without a maximum or a minimum you can't use a sign table to determine where the function decreases or increases. So I used the graph of the first derivative. Why is it not correct?
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  5. #5
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    Re: Graphing using calculus

    Hello Guys.
    What I did for part c and d was solve 1 +2xarctan(x) = 0

    which i found to be arctan(x) = -1/2x

    since arctan is > -1/2x on (0, inf) and < -1/2x on (-inf,0) i figured that would be the points when k was increasing and decreasing respectively
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  6. #6
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    Re: Graphing using calculus

    Quote Originally Posted by andrew2322 View Post
    Hello Guys.
    What I did for part c and d was solve 1 +2xarctan(x) = 0

    which i found to be arctan(x) = -1/2x

    since arctan is > -1/2x on (0, inf) and < -1/2x on (-inf,0) i figured that would be the points when k was increasing and decreasing respectively
    1 + 2x\arctan{x} \ge 1 for all x ...
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  7. #7
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    Re: Graphing using calculus

    Quote Originally Posted by skeeter View Post
    1 + 2x\arctan{x} \ge 1 for all x ...
    I dont understand?
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  8. #8
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    Re: Graphing using calculus

    \arctan{x} =0 only at x = 0.

    So that 2x\arctan{x}=0 only at x = 0, and 2x\arctan{x} is an even function with an absolute minimum of 0 at x = 0.

    Therefore, 2x\arctan{x}\ge 0\,.

    This gives you 1+2x\arctan{x}\ge 1\,.

    The first derivative is never zero.
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