Calculus
posted by anonymous .
Consider the function f(x)=xe^2x. Find the intervals on which f(x) is concave up and the intervals where it is concave down. Show a sign graph.
Please show work in detail so I can follow for future reference. Thanks!

The curve is concave up if f '' (x) is positive and
concave down if f '' (x) is negative
f ' (x) = x(2)e^(2x) + e^(2x)
= e^(2x) (2x + 1)
f '' (x) = (2x+1)(2)e^(2x) + 2 e^(2x)
= 2 e^(2x) ((2x+1) + 1)
= 2 e^(2x) (2x + 2)
= 4 e^(2x) (x+1)
since 4e^(2x) > 0 for all values of x, we just have to look at
x + 1
x+1 > 0 for x > 1
x+1 < 0 for x < 1
so the curve is concave up for all values of x > 1
and concave down for all values of x < 1
check the graph by Wolfram
http://www.wolframalpha.com/input/?i=plot+x+e%5E%282x%29%2C+3+%3C+x+%3C+1+%2C+.25+%3C+y+%3C+2
notice there is a point of inflection at x= 1