For what values of x is f(x)=-x^2+e^xf(x)=x2+ex concave or convex?

1 Answer
Mar 25, 2018

(-oo, ln2):(,ln2): Convex (ln2, oo): Concave(ln2,):Concave

Explanation:

Determining concavity requires finding f''(x) and examining where it is positive or negative.

f'(x)=-2x+e^x

f''(x)=-2+e^x

To determine where this is positive or negative, let's set f''(x)=0 and solve.

-2+e^x=0

e^x=2

ln(e^x)=ln(2)

x=ln(2)

Let's split up the domain of f(x), (-oo, oo), around x=ln(2), and test values for f''(x) to see if it is positive or negative:

(-oo, ln(2)): f''(0)=-2+e^0=-1<0

Thus, since f''(x)<0 on (-oo, ln(2)), f(x) is convex on (-oo, ln(2)).

(ln(2), oo):

f''(ln(3))=-2+e^ln3=-2+3>0

Thus, since f''(x)>0 on (ln(2), oo), f(x) is concave on (ln(2), oo).