Asked by Rene
Find any relative extrema and any points of inflection if they exist of f(x)=x^2+ ln x^2 showing calculus. Please show work in detail so I can follow. Thanks. The answer is no relative max or min and the points of inflection are (1,1) and (-1,1) but I'm having trouble getting there
Answers
Answered by
Steve
f(x) = x^2 + lnx^2
f' = 2x + 2/x = 2(x^2+1)/x
f" = 2 - 2/x^2 = 2(x^2-1)/x^2
It's clear f' is never 0, so no min/max
f"=0 when x=1 or -1
so, the points of inflection are (-1,1) and (1,1)
Note that ln(x^2) is defined for all x not zero, since x^2>=0. So its domain includes -1, while 2ln(x) does not, even though we can usually simplify ln(x^2) = 2ln(x).
f' = 2x + 2/x = 2(x^2+1)/x
f" = 2 - 2/x^2 = 2(x^2-1)/x^2
It's clear f' is never 0, so no min/max
f"=0 when x=1 or -1
so, the points of inflection are (-1,1) and (1,1)
Note that ln(x^2) is defined for all x not zero, since x^2>=0. So its domain includes -1, while 2ln(x) does not, even though we can usually simplify ln(x^2) = 2ln(x).
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