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simple problem


razorfane

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i got .26595745 and 1.4361702.

i'm not sure about that' date=' but the gradient is [math']y=x^{2x}-2x\rightarrow\frac{dy}{dx}=2x^{2x+1}-2[/math].

 

I just had to point out that the above is not correct. The derivative of [math]x^{2x}[/math] is [math]2(ln(x)+1)x^{2x}[/math]

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i solved it graphically. i used my calculator this time and got closer values than before. they are .25 and aproxamately 1.443737. i set it =0 and assumed y=0. my answers are where the function crosses the x-axis.

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Oh, ok. Actually there is no algebraic solution and... i'm joking, here it is:

 

[math]x^{2x}=2x[/math]

[math]x^{2x-1}=2[/math]

[math]x^{\frac{2x-1}{2}}=2^{\frac{1}{2}}[/math]

[math]x^{x-\frac{1}{2}}=2^{\frac{1}{2}}[/math]

[math]x^{x-\frac{1}{2}}=\left(\tfrac{1}{2}\right)^{-\frac{1}{2}}[/math]

[math]x^{x-\frac{1}{2}}=\left(\tfrac{1}{4}\right)^{-\frac{1}{4}}[/math]

[math]x^{x-\frac{1}{2}}=\left(\tfrac{1}{4}\right)^{\frac{1}{4}-\frac{1}{2}}[/math]

Since both members are structurally the same, you deduct:

[math]X=\frac{1}{4}[/math]

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okay, I understand everything up until this point.

[math]x^{x-\frac{1}{2}}=\left(\tfrac{1}{4}\right)^{\frac{1}{4}-\frac{1}{2}}[/math]

 

Will you please explain how you got from the previous step to that one? Also, what about the other answer, 1.4437374? Is there a way to get that one?

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