f' = x (lnx)^2
f" = lnx(lnx+2)
f is concave down where f" < 0, or
-2 < lnx < 0
That is, on the interval (1/e^2,1)
You can see the graph of f(x) at
http://www.wolframalpha.com/input/?i=1%2F4+x^2+%282%28lnx%29^2-2lnx%2B1%29+for+0%3Cx%3C2
Let f(x) be a function such that f′(x)=x(ln|x|)^2.
Find the open interval(s) in which f(x) is concave down .
In interval notation, f(x) is concave down in ___________
1 answer
1 answer