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First one:
f'(x) = 2x*lnx + x^2*(1/x) = 2x*lnx + x = x*(2*lnx + 1)
Second derivate:
f"(x) = D [x*(2*lnx + 1)] = 1*(2*lnx + 1) + x*(2/x)
= 2*lnx+1+2
= 2*lnx + 3
So, there is a minimum in this graph on point (1/e^(1/2)), -1/(2e)) = MIN(x, y)
Van Sanden David
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∙ 11y ago
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