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Thread: Prove the radical in the Equation

  1. #1
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    Prove the radical in the Equation

    be cosx=x^2.

    Prove that this equation has exactly two radical .
    how i do it?

    thanks for help
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  2. #2
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    Quote Originally Posted by ZOOZ View Post
    be cosx=x^2.
    Prove that this equation has exactly two radical .
    I think that you mean 'has exactly two roots'.
    Look at the function $\displaystyle f(x)=x^2-\cos(x)$.
    Show that it has two zeros in $\displaystyle [-1,1]$.
    Then show its derivative is negative for numbers less than zero and positive for numbers greater than zero.
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  3. #3
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    well cosx has two zeros in [-pi/2,pi/2] but i still no understand how to prove this function has exactly two roots maybe i need to use in Rolle's theorem can u Please show me how can i prove it?

    thanks.
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  4. #4
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    The zeros of cos(x) are irrelevant here. As Plato suggested, look at $\displaystyle x^2- cos(x)$ on the interval [-1, 1]. $\displaystyle 1- cos(1)> 0$, $\displaystyle 0^2- cos(0)< 0$, and $\displaystyle 1^2- cos(1)> 0$.

    The derivative of $\displaystyle cos(x)- x^2$ is $\displaystyle -sin(x)- 2x$. If you can show that derivative is negative for x> 0, as Plato suggested, then you know $\displaystyle cos(x)- x^2$ is "one-to-one" and has at most one zero for x>0. Similarly for x< 0.
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  5. #5
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    well ok for f'=-sin(x)-2x the derivative is negative for x>0

    and for f'=-sinx-2x the derivative is Positive for x<0
    this i Understand but when u told "then you know cos(x)-x^2 is "one-to-one" and has at most one zero for x>0. Similarly for x< 0."

    for what i need all the part A of ur answer i just can use the part with the derivative no?
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  6. #6
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    Quote Originally Posted by ZOOZ View Post
    well ok for f'=-sin(x)-2x the derivative is negative for x>0 and for f'=-sinx-2x the derivative is Positive for x<0
    this i Understand but when u told "then you know cos(x)-x^2 is "one-to-one" and has at most one zero for x>0. Similarly for x< 0.
    That post is practically unintelligible. Please practice using standard words, complete sentences and NO web-speak as ‘ur’.

    $\displaystyle f(-1)=1-\cos(-1)>0,~f(0)=-1<0,~\&~f(1)>0$ that means it has a root between $\displaystyle -1~\&~0$ and a root between $\displaystyle 0~\&~1$.
    The derivative is negative for negative values of $\displaystyle x$ so there is one negative root.
    Likewise there is one positive root.
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  7. #7
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    Mean value theorem: If f is differentiable on [a, b] then there exist a number c, in [a, b], such that $\displaystyle \frac{f(b)- f(a)}{b- a}= f'(0)$. It there were more than one root of the equation f(x)= 0 we could take a and b to be those roots and have $\displaystyle \frac{f(b)- f(a)}{b- a}= 0 = f'(c)$. That is, if there is more than one root, there would have to be a point beteen them where f'(c)= 0. But f'(x) here is 0 only at x= 0. There cannot be two positive or two negative value of x such that $\displaystyle cos(x)- x^2= 0$.
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  8. #8
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    hi thank you for the asnwers but you have worng in 1 time you told to look at X^2-cosx on the interval [-1, 1] that ok.
    BUT then you told The derivative of cos(x)-x^2.


    so the derivative of cos(x)-x^2 is not like the derivative of X^2-cosx.
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  9. #9
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    well ok i Understand how to Solve it now.

    Thanks to you'r help.

    so i really Grateful about all the help you gave me, all of you.
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