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If you know differentiation, then you can easily calculate it by just setting it's derivative equal to 0. Another method is just by looking at it's graph from a graphing calculator.
Let y=xx=exlnx. Then dxdy=(lnx+1)exlnx. Since exlnxxx>0, dxdy=0, when lnx+1=0 or lnx=−1⟹x=e−1=e1. Note that dx2d2y=xxx+(lnx+1)2xx>0, when x=e1, implies that y=xx is minimum when x=e1 and min(xx)=e−e−1≈0.692.
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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
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[example link](https://brilliant.org)> This is a quote# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"\(...\)or\[...\]to ensure proper formatting.2 \times 32^{34}a_{i-1}\frac{2}{3}\sqrt{2}\sum_{i=1}^3\sin \theta\boxed{123}Comments
If you know differentiation, then you can easily calculate it by just setting it's derivative equal to 0. Another method is just by looking at it's graph from a graphing calculator.
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Do you mean calculate a point where the slope is zero ?
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Yes,exactly. That will give you the minima because the function doesn't has any maxima (it's an increasing function).
Hint: Implicit differentiation.
Let y=xx=exlnx. Then dxdy=(lnx+1)exlnx. Since exlnxxx>0, dxdy=0, when lnx+1=0 or lnx=−1 ⟹x=e−1=e1. Note that dx2d2y=xxx+(lnx+1)2xx>0, when x=e1, implies that y=xx is minimum when x=e1 and min(xx)=e−e−1≈0.692.