Posted by **anonymous** on Saturday, April 20, 2013 at 9:42pm.

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!

- Calculus -
**Reiny**, Saturday, April 20, 2013 at 10:33pm
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

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