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Math Help - minimum sign question

  1. #1
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    minimum sign question

    f(x)=x^{4}+(sinx)e^{-x^{2}}
    prove that if x_{0} is the minimum point
    then f(x_{0})<0
    hint: look at f(0) and f'(0).
    how i tried to solve it.
    f(0)=0 f'(0)=1
    by limit derivative definition f'(0)=lim_{x->0}\frac{f(0+x)-f(0)}{x-0}=lim_{x->0}\frac{f(x)}{x}=1
    so by limit definition for -\delta<x<\delta
    1-\epsilon<\frac{f(x)}{x}<1+\epsilon
    what now?
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  2. #2
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    Re: minimum sign question

    If f(0) = 0, and f'(0) = 1, then what must happen just to the left of the origin?
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  3. #3
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    Re: minimum sign question

    what it has to do with how i tried to solve it?
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    Re: minimum sign question

    If you suspect that f(x) < 0 for some x to the left of the origin, then posit epsilon less than 1 in magnitude, and x < 0. What does that do to your inequalities?
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  5. #5
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    Re: minimum sign question

    we are talking about the value on x0
    why you say " for some x to the left of the origin"
    why you say that x0<0
    ?
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  6. #6
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    Re: minimum sign question

    Quote Originally Posted by transgalactic View Post
    we are talking about the value on x0
    why you say " for some x to the left of the origin"
    why you say that x0<0
    ?
    Think about Ackbeet's suggestion: if f(0)= 0 and f'(0)> 0 (so that f is an increasing function there), what can you say about f(x) for x slightly less than 0?
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  7. #7
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    Re: minimum sign question

    Quote Originally Posted by transgalactic View Post
    we are talking about the value on x0
    why you say " for some x to the left of the origin"
    why you say that x0<0
    ?
    Because it's not too hard to show that f(x) > 0 for all x > 0. Think about it. Everything in sight is positive until x = pi. At that point, the x^4 completely dwarfs the sin(x), which is modulated by something that's getting smaller anyway.

    Plot the function, and you'll see what I mean. The minimum happens to the left of the origin, right around -0.46.
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  8. #8
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    Re: minimum sign question

    yes i understand that for x<0 the function is negative
    and that our minimum has to be less then f(0)=0 .

    ok but its only words i have to show that in the equation.
    if i choose epsilon =1/2

    so by limit definition for -\delta<x<\delta 0.5<\frac{f(x)}{x}<1.5
    what to do next
    ?
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  9. #9
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    Re: minimum sign question

    Take the inequality in the OP, and multiply through by x, which is negative. What happens to the inequality signs? And then what can you say?
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  10. #10
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    Re: minimum sign question

    if we multply by negatibe x
    so by limit definition for -\delta<x<\delta 0.5x>f(x)>1.5x
    but i cant see how from that we can conclude that the minimum is negative


    i can guess that
    for every negative x
    f(x) is between two negative values
    so it have to be negative by the intemideate value theorem?

    but i dont have actual points the for the intermidieate value theorem
    i have expressions of x


    or does it sufficient to say that its between two negatives so its negative too.

    is it a proper proof
    ?
    Last edited by transgalactic; September 15th 2011 at 12:05 PM.
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  11. #11
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    Re: minimum sign question

    Quote Originally Posted by transgalactic View Post
    if we multply by negatibe x
    so by limit definition for -\delta<x<\delta 0.5x>f(x)>1.5x
    but i cant see how from that we can conclude that the minimum is negative
    0>0.5x>f(x)>1.5x.

    i can guess that
    for every negative x
    f(x) is between two negative values
    so it have to be negative by the intemideate value theorem?

    but i dont have actual points the for the intermidieate value theorem
    i have expressions of x
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  12. #12
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    Re: minimum sign question

    but our surriundings is -delta<x<0

    not delta from both sides
    correct?
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  13. #13
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    Re: minimum sign question

    Look. You have

    1-\epsilon<\frac{f(x)}{x}<1+\epsilon.

    If you choose \epsilon=0.5, then

    0.5<\frac{f(x)}{x}<1.5.

    Multiplying by an x that we assume is negative yields

    0.5x>f(x)>1.5x.

    But, since x<0, it follows that 0.5x<0. Hence, we can tack on the inequality to our previous one thus:

    0>0.5x>f(x)>1.5x.
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  14. #14
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    Re: minimum sign question

    yes but in the limit definition whe have for
    -delta<x<delta

    but we talk only about the negative x
    so only half of the section is being uses

    so in the limit definition what section two write
    ?
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  15. #15
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    Re: minimum sign question

    So the logic works like this: since the limit exists and is 1, it is the case that for every \epsilon>0, there is a \delta>0 such that if |x|<\delta, then

    \left|\frac{f(x)-f(0)}{x-0}-1\right|<\epsilon.

    Let \epsilon=0.5. Then there is a \delta_{0.5}>0 such that if |x|<\delta_{0.5}, then |(f(x)/x)-1|<0.5.

    Pick x_{1}\in(-\delta_{0.5},0). Then |x_{1}|<\delta_{0.5}. It follows that

    |(f(x_{1})/x_{1})-1|<0.5.

    And then you work out the inequalities just like before, in my post # 13, only with x_{1}'s instead of x's. Finally, you can say that

    f(x_{0})<f(x_{1}), since x_{0} is the minimum.
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