Question 198715
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ g(x) = y = \frac{x^x}{4}]


Obtain the first derivitive:


First take the natural log of both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln(y) = \ln\left(\frac{x^x}{4}\right) = x\ln(x) - \ln(4)]


Differentiate with the chain rule on the left and the product rule on the right:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dy}{dx}\left(\frac{1}{y}\right)\ =\ \ln(x) + x\left(\frac{1}{x}\right) + 0\ =\ \ln(x) + 1]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dy}{dx}\ =\ \left(\ln(x) + 1\right)y\ = \ \left(\ln(x) + 1\right)x^x]


Set the first derivative equal to zero:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\ln(x) + 1\right)x^x = 0]


Since *[tex \Large x^x > 0\ \forall\ x > 0],


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\ln(x) + 1\right)x^x = 0] if and only if


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln(x) + 1 = 0]


Hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln(x) = -1 \ \ \Rightarrow\ \ x = e^{-1} \approx 0.37]


Therefore *[tex \Large \left(e^{-1},g\left(e^{-1}\right)\right)] is a local extreme point.


The calculator tells us that *[tex \Large g\left(e^{-1}\right) \approx .17] but *[tex \Large 0 < e^{-1} < 1], *[tex \Large g\left(0\right) = .25] and *[tex \Large g\left(1\right) = .25], so we can be assured that this is a minimum.






John
*[tex \LARGE e^{i\pi} + 1 = 0]
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