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Math Help - solving equations

  1. #1
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    solving equations

    I was looking for critical points and found my f'(x)=-12/(4x^2+1)(\sqrt{4x^2+1})

    Now I want to solve that equation , by setting
    -12/(4x^2+1)(\sqrt{4x^2+1}=0

    will there be any solution for x? or this is an equation that has no solution?
    Last edited by ameerulislam; April 18th 2013 at 06:41 AM.
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  2. #2
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    Re: solving equations

    Quote Originally Posted by ameerulislam View Post
    I was looking for critical points and found my f'(x)=-12/(4x^2+1)(\sqrt{4x^2+1})

    Now I want to solve that equation , by setting
    -12/(4x^2+1)(\sqrt{4x^2+1}=0

    will there be any solution for x? or this is an equation that has no solution?
    Is it that equation that you meant to write?

    f'(x)= \frac{-12\sqrt{4x^2+1}}{4x^2+1}
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  3. #3
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    Re: solving equations

    Quote Originally Posted by dokrbb View Post
    Is it that equation that you meant to write?

    f'(x)= \frac{-12\sqrt{4x^2+1}}{4x^2+1}
    No, only -12 is up there.

    f'(x)= \frac{-12}{(\sqrt{4x^2+1})(4x^2+1)}
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  4. #4
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    Re: solving equations

    so

    what would be the solution for

     \frac{-12}{(\sqrt{4x^2+1})(4x^2+1)} =0
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  5. #5
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    Re: solving equations

    Quote Originally Posted by ameerulislam View Post
    so

    what would be the solution for

     \frac{-12}{(\sqrt{4x^2+1})(4x^2+1)} =0
    I'm not sure, but working with the denominator ...

     \frac{-12}{({4x^2+1}^{\frac{1}{2}})(4x^2+1)^{1}}=0

    it gives you this  \frac{-12}{{4x^2+1}^{\frac{3}{2}}}=0 you have a constant at the numerator, and apparently the denominator is never 0 because of
    {4x^2+1}^{\frac{3}{2}}}

    could you provide as with the original  f(x) ?

    dokrbb
    Last edited by dokrbb; April 18th 2013 at 07:20 AM.
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  6. #6
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    Re: solving equations

    sure, this was the question

    f(x)= 3/sqrt(4x^2 + 1) ; [-1,1] , the question asked to find the absolute max and min.
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  7. #7
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    Re: solving equations

    you were supposed to get this as derivative  \frac{-12x}{{4(x)^2+1}^{\frac{3}{2}}}

    and because of -12x the only critical point would be at x=0

    can you solve for x now
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  8. #8
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    Re: solving equations

    Quote Originally Posted by dokrbb View Post
    you were supposed to get this as derivative  \frac{-12x}{{4(x)^2+1}^{\frac{3}{2}}}

    and because of -12x the only critical point would be at x=0

    can you solve for x now
    oh.. you are right...


    Ok now we have 3 candidates for relative maxima


    one CP: 0,
    and 2 End points: 1 and -1


    if I plug those numbers on the original equation


    \frac{3}{\sqrt{(4x^2 + 1)}}

    i get

    For 0 , I get 3
    for -1, I get \frac{3}{\sqrt{5}}

    for 1, I get \frac{3}{\sqrt{5}}

    Am I right? so absolute maximum is 0 and absolute minimum is -1 and 1 ?

    But in my book (in the answer section) it says something else
    maximum value \frac{3}{\sqrt{5}} at x=1

    minimum value -\frac{3}{\sqrt{5}} at x=-1

    I have no idea how did they get -\frac{3}{\sqrt{5}} , that x was squared right ?
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  9. #9
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    Re: solving equations

    Quote Originally Posted by ameerulislam View Post
    oh.. you are right...


    Ok now we have 3 candidates for relative maxima


    one CP: 0,
    and 2 End points: 1 and -1


    if I plug those numbers on the original equation


    \frac{3}{\sqrt{(4x^2 + 1)}}

    i get

    For 0 , I get 3
    for -1, I get \frac{3}{\sqrt{5}}

    for 1, I get \frac{3}{\sqrt{5}}

    Am I right? so absolute maximum is 0 and absolute minimum is -1 and 1 ?

    But in my book (in the answer section) it says something else
    maximum value \frac{3}{\sqrt{5}} at x=1

    minimum value -\frac{3}{\sqrt{5}} at x=-1

    I have no idea how did they get -\frac{3}{\sqrt{5}} , that x was squared right ?
    I'm sure someone would correct me on this point but since your function is defined in the interval [-1, 1] the value of x=0 doesn't respect the conditions for the absolute maxima and minima, and in this case you woul like to look for maxima and minima at the endpoints.

    Here is a more or less detailed description - "Restricted domains: There may be maxima and minima for a function whose domain does not include all real numbers. A real-valued function, whose domain is any set, can have a global maximum and minimum. There may also be local maxima and local minima points, but only at points of the domain set where the concept of neighborhood is defined. A neighborhood plays the role of the set of x such that |x − x∗| < ε."

    dokrbb
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  10. #10
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    Re: solving equations

    I just used and online math calculator and I feel my observation suppose to be correct

    solving equations-math_problem_masima.jpg
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  11. #11
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    Re: solving equations

    Quote Originally Posted by ameerulislam View Post
    I just used and online math calculator and I feel my observation suppose to be correct

    Click image for larger version. 

Name:	math_problem_masima.JPG 
Views:	4 
Size:	47.5 KB 
ID:	28008
    Yes, but this is not for the function which is restricted on [-1, 1] interval - that's what makes the difference in your case,

    see that as an example Maxima and minima - Wikipedia, the free encyclopedia

    dokrbb
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  12. #12
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    Re: solving equations

    Your argument seems to be correct, and the graph makes it pretty obvious. There's a maximum of 3 at x=0 and minima of \frac{3}{\sqrt{5}} at x=1 and x=-1. Since the function is restricted to -1 \le x \le 1, you ignore the parts of the graph to the right of x=1 and to the left of x=-1.

    It would seem your book is wrong. Assuming you've relayed the question and solution correctly........

    - Hollywood
    Thanks from MarkFL
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  13. #13
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    Re: solving equations

    Sorry guys.. I made another mistake here... Just got crazy before the exam :/...

    The original equation had an extra x at the top..

    f(x)= 3x/sqrt(4x^2 + 1); [-1,1]

    grrrrr...
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  14. #14
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    Re: solving equations

    Quote Originally Posted by ameerulislam View Post
    Sorry guys.. I made another mistake here... Just got crazy before the exam :/...

    The original equation had an extra x at the top..

    f(x)= 3x/sqrt(4x^2 + 1); [-1,1]

    grrrrr...
    I would show your graph now - click here 3x/sqrt(4x^2 + 1) - Wolfram|Alpha

    as you see, your function is continuous at 0, and there is actually no maxima or minima, but since you have the restricted domain - there it is,

    Next time just post the whole problem, not just the parts you may think a re needed/important/relevant, would be easier for you and for all another that are willing to understand the problems here,

    dokrbb
    Thanks from ameerulislam
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  15. #15
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    Re: solving equations

    Quote Originally Posted by dokrbb View Post
    I would show your graph now - click here 3x/sqrt(4x^2 + 1) - Wolfram|Alpha

    as you see, your function is continuous at 0, and there is actually no maxima or minima, but since you have the restricted domain - there it is,

    Next time just post the whole problem, not just the parts you may think a re needed/important/relevant, would be easier for you and for all another that are willing to understand the problems here,

    dokrbb
    Wow, that site is amazingly crazy!!!
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