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Math Help - [SOLVED] Family of functions

  1. #1
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    [SOLVED] Family of functions

    Given the family of functions h(x)=x+cos(ax), where a is a positive constant such that 0<a<4.

    For what values of a wiill h(x) have a relative maximum at x=1?

    I found the h'(x) first. Then I equated that to zero: h'(x)=1-asin(ax).
    Then I'm not sure what to do next, since there are two variables, do I just plug x=1 into the equation then solve for a?
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    Quote Originally Posted by chukie View Post
    Given the family of functions h(x)=x+cos(ax), where a is a positive constant such that 0<a<4.

    For what values of a wiill h(x) have a relative maximum at x=1?

    I found the h'(x) first. Then I equated that to zero: h'(x)=1-asin(ax).
    Then I'm not sure what to do next, since there are two variables, do I just plug x=1 into the equation then solve for a?
    Yes.

    But remember that the solutions you get for a force stationary points at x = 1 and so might include values that give relative minimum and stationary points of inflexion at x = 1 .... You need to test the nature of the solutions .....

    Also, you won't be able to solve \sin a = \frac{1}{a} for exact values of a so you'll have to content yourself with approximate solutions ......
    Last edited by mr fantastic; July 2nd 2008 at 08:20 PM. Reason: Added italics on a.
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    Quote Originally Posted by mr fantastic View Post
    Yes.

    But remember that the solutions you get for a force stationary points at x = 1 and so might include values that give relative minimum and stationary points of inflexion at x = 1 .... You need to test the nature of the solutions .....

    Also, you won't be able to solve \sin a = \frac{1}{a} for exact values of a so you'll have to content yourself with approximate solutions ......
    Thanks! I've managed to find a by graphing it on my calculator. Now the next question asks me what values of a will h(x) have an inflection pt at x=1. I have found the second derivative for h(x), equated that to zero, then plug 1 in again to solve for a. I graphed it again to find the values for a, except there are no zeroes when I graphed it. Am I doing something wrong?
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    Quote Originally Posted by chukie View Post
    Thanks! I've managed to find a by graphing it on my calculator. Now the next question asks me what values of a will h(x) have an inflection pt at x=1. I have found the second derivative for h(x), equated that to zero, then plug 1 in again to solve for a. I graphed it again to find the values for a, except there are no zeroes when I graphed it. Am I doing something wrong?
    h''(x) = -a^2 \cos (ax). The equation 0 = -a^2 \cos a has plenty of solutions.

    You should be able to get exact solutions for a without using technology.

    Note: h''(1) = 0 is a necessary but not sufficient condition for an inflexion point at x = 1. So you'll need to test all solutions to this equation.
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    Quote Originally Posted by mr fantastic View Post
    h''(x) = -a^2 \cos (ax). The equation 0 = -a^2 \cos a has plenty of solutions.

    You should be able to get exact solutions for a without using technology.

    Note: h''(1) = 0 is a necessary but not sufficient condition for an inflexion point at x = 1. So you'll need to test all solutions to this equation.
    Ah! I see, I did the second derivative wrong. One last question, what values of a make h(x) strictly decreasing.

    I was thinking that if it is strictly decreasing, the function would be monotonic, so h'(x)<0. Thus I wrote 1-asin(ax)<0, but now what value of x should I plug in? Is it still 1?
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    Quote Originally Posted by chukie View Post
    Ah! I see, I did the second derivative wrong. One last question, what values of a make h(x) strictly decreasing.

    I was thinking that if it is strictly decreasing, the function would be monotonic, so h'(x)<0. Thus I wrote 1-asin(ax)<0, but now what value of x should I plug in? Is it still 1?
    If the question is interpretted as h(x) is decreasing at x = 1, then you require h'(1) < 0 \Rightarrow 1 - a \sin a < 0.
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    Quote Originally Posted by mr fantastic View Post
    If the question is interpretted as h(x) is decreasing at x = 1, then you require h'(1) < 0 \Rightarrow 1 - a \sin a < 0.
    The question doesn't say that though; it just says which values of a makes h(x) strictly decreasing. If I don't interpret it as h(x) is decreasing at x = 1, can the question be done?

    I was trying out different graphs on my calculator with a. I have noticed that when 0<a<1 the graph seems to be monotonic, but increasing. I don't think there is a value for a that makes h(x) decreasing. But this is just a guess.
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    Quote Originally Posted by chukie View Post
    The question doesn't say that though; it just says which values of a makes h(x) strictly decreasing. If I don't interpret it as h(x) is decreasing at x = 1, can the question be done?

    I was trying out different graphs on my calculator with a. I have noticed that when 0<a<1 the graph seems to be monotonic, but increasing. I don't think there is a value for a that makes h(x) decreasing. But this is just a guess.
    A necessary condition is that y = h(x) does not have any stationary points => values of a such that h'(x) \neq 0 \Rightarrow 1 - a \sin (ax) \neq 0 \Rightarrow a \sin (ax) \neq 1 for all x. So a necessary condition is that 0 < a < 1.

    However, this could mean that y = h(x) is decreasing OR increasing ......


    y = h(x) is decreasing => values of a such that h'(x) < 0 \Rightarrow 1 - a \sin (ax) < 0 \Rightarrow a \sin (ax) > 1 for all x.

    But it's impossible to satisfy this condition AND a \sin (ax) \neq 1.

    So the answer is that there's no values of a for which h(x) is decreasing for all values of x. Does the question give an interval over which h(x) is to be strictly decreasing?
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    Quote Originally Posted by mr fantastic View Post
    Does the question give an interval over which h(x) is to be strictly decreasing?
    Nope.
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    Quote Originally Posted by chukie View Post
    Nope.
    Then there's no value of a.
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    Awesome! Thanks so much for your help!
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