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

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Answer 1 · 2018-03-25T20:19:44

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

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)#.