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Where does the second derivative of (x^2)/[(x^2)+3] = 0?
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Using the quotient rule, which you should know we get that.... The first derivative of that is: \[\frac{ 6x }{(3+x^2)^2 }\] The derivative of that is: \[\frac{-(18(-1+x^2)) }{ (3+x^2)^3 }\] Once we simplify the above fraction which you should be able to do we get: \[x^2-1=0\] Which is at x=1 and x=-1. Make sense?
Ah yes, I was making a stupid sign mistake that I did not catch until I saw your second derivative. Thank you!
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