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Math Help - solving equation

  1. #1
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    solving equation

    Hi. I have a maths question on finding the turning points and determining their nature but I can't find the value of x and w.

    The equation is z= e^(2x) + e^(-y) + e^(w^2) - e^(ln2x) - 2e^(w) + y

    I differentiated it wrt to x, y and w.

    But I can't solve for x and w in the following equations:

    2e^(2x) - [ (e^ln2x) / x) = 0

    2we^(w^2) - 2e^(w) = 0


    I really need help for this one. Thanks.
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  2. #2
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    \displaystyle 2e^{2x} - \frac{e^{\ln{(2x)}}}{x} = 0

    \displaystyle 2e^{2x} - \frac{2x}{x} = 0

    \displaystyle 2e^{2x} - 2 = 0

    \displaystyle 2e^{2x} = 2

    \displaystyle e^{2x} = 1

    \displaystyle 2x = 0

    \displaystyle x = 0.


    \displaystyle 2w\,e^{w^2} - 2e^w = 0

    Simple observation will show that \displaystyle w = 1 is a solution.
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  3. #3
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    I did what you did but it's not right. x cannot be equal to 0, otherwise if you replace it in 2e^(2x) - [ (e^ln2x) / x) = 0,
    (e^ln2x) / 0 will be undefined and I need a value of x. And for the second equation how would I know if w has only one value and not 2?
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  4. #4
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    Quote Originally Posted by nepmma View Post
    I did what you did but it's not right. x cannot be equal to 0, otherwise if you replace it in 2e^(2x) - [ (e^ln2x) / x) = 0,
    (e^ln2x) / 0 will be undefined and I need a value of x. And for the second equation how would I know if w has only one value and not 2?
    Yes, you are right, \displaystyle x \neq 0 if you leave the function written as is. You can simplify your original function though, because \displaystyle e^{\ln{(2x)}} = 2x. Otherwise the function will be not be defined for any nonpositive value anyway...


    In the second case, you would need to show that \displaystyle 2w\,e^{w^2} - 2e^w does not have any turning points.
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    Quote Originally Posted by Prove It View Post
    Yes, you are right, \displaystyle x \neq 0 if you leave the function written as is. You can simplify your original function though, because \displaystyle e^{\ln{(2x)}} = 2x. Otherwise the function will be not be defined for any nonpositive value anyway...


    In the second case, you would need to show that \displaystyle 2w\,e^{w^2} - 2e^w does not have any turning points.


    Ohhh!! I forgot the basic laws of logarithms..I forgot I could simplify it(I feel stupid now). Thanks!!! But how am I supposed to show that that 2we^(w^2) - 2e^(w) is always increasing and doesn't have any turning pts? Do I have to plot a graph or something?
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    Quote Originally Posted by nepmma View Post
    Ohhh!! I forgot the basic laws of logarithms..I forgot I could simplify it(I feel stupid now). Thanks!!! But how am I supposed to show that that 2we^(w^2) - 2e^(w) is always increasing and doesn't have any turning pts? Do I have to plot a graph or something?
    How do you usually find the turning point of a function? Find the derivative, set it equal to zero, see if there are any solutions. Then apply the second derivative test.
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