solving equation

**nepmma** · November 10th 2010, 01:24 PM

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.

**Prove It** · November 10th 2010, 01:32 PM

$\displaystyle 2e^{2x} - \frac{e^{\ln{(2x)}}}{x} = 0$

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

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

$\displaystyle 2e^{2x} = 2$

$\displaystyle e^{2x} = 1$

$\displaystyle 2x = 0$

$\displaystyle x = 0$ .

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

Simple observation will show that $\displaystyle w = 1$ is a solution.

**nepmma** · November 10th 2010, 01:37 PM

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?

**Prove It** · November 10th 2010, 01:47 PM

Originally Posted by nepmma

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.

**nepmma** · November 10th 2010, 02:00 PM

Originally Posted by Prove It

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?

**Prove It** · November 10th 2010, 02:49 PM

Originally Posted by nepmma

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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