## Sarperson Group Title Can you help me find the critical points of this equation? h(x)=x^2lnx^2 one year ago one year ago

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1. SomeBloke Group Title

Yes.

2. estebananaya Group Title

i think h(x) has only one asymptote on x=0 and has 1 critical point at x=e^-1/2

3. estebananaya Group Title

actually, i think there are two critical points, one at x=e^1/2 and another at x=-e^1/2

4. crimsonblackarc Group Title

First solve for h'(x) $h\prime(x)=x^{2}(\frac{ 2x }{ x^{2}})+2xln(x^{2})$ $h \prime(x) =2x(1+\ln(x ^{2}))$ Equating h'(x)=0 will give you the first critical point, which is x=0. We can find the other two by equating 1+ln(x^2)=0 $-1=\ln (x ^{2})$ $e ^{-1}=x^2$ $\frac{ 1 }{e }=x^2$ $x= \pm \sqrt{\frac{ 1 }{ e }}$