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Answer: 1/e


The turning point of x^x is the point where the derivative of x^x is 0.

To find the derivative of x^x, let y=x^x. Thus, ln(y)=ln(x^x), so ln(y)=x*ln(x). Now, we differentiate.

Using the chain rule on the left side gives (1/y)*y'. Using the product rule on the right side gives 1*ln(x)+x*(1/x), which simplifies to ln(x)+1.

Thus, we have (1/y)*y'=ln(x)+1. Since we are looking for y', multiply both sides by y to give y'=(ln(x)+1)y. Since y=x^x, the derivative of x^x is (ln(x)+1)(x^x).

Now, we need to make (ln(x)+1)(x^x)=0. For this to happen, either ln(x)+1=0 or x^x-0. x^x will never equal 0 because, if it did, you could put both sides to the power 1/x and you would get x=0 unless x=0(in which case 1/x is undefined, so you could not put both sides to that power), and 0^0 is undefined (and therefore not 0).

Therefore, the turning point is where ln(x)+1=0. Thus, ln(x)=-1. By definition, this means that e^-1=0. Again, by definition, x=1/e.

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Q: What is the turning point of the graph x to the power of x?
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