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estebananaya
 2 years ago
Best ResponseYou've already chosen the best response.0i think h(x) has only one asymptote on x=0 and has 1 critical point at x=e^1/2

estebananaya
 2 years ago
Best ResponseYou've already chosen the best response.0actually, i think there are two critical points, one at x=e^1/2 and another at x=e^1/2

crimsonblackarc
 2 years ago
Best ResponseYou've already chosen the best response.0First 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 \[1=\ln (x ^{2})\] \[e ^{1}=x^2\] \[\frac{ 1 }{e }=x^2\] \[x= \pm \sqrt{\frac{ 1 }{ e }}\]
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