For what values of x is #f(x)=-x^2+4x+2# concave or convex?

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Answer 1 · 2016-12-03T07:21:28

The function is concave.

#f(x)=-x^2+4x+2#

#f'(x)=-2x+4#

For critical points, #f'(x)=0#

#-2x+4=0#

#x=2#

Let's do a sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##2##color(white)(aaaa)##+oo#

#color(white)(aaaa)##f'(x)##color(white)(aaaa)##+##color(white)(aa)##0##color(white)(aaa)##-#

#color(white)(aaaa)##f(x)##color(white)(aaaaa)##uarr##color(white)(aa)##6##color(white)(aaa)##darr#

So #f(x)# is concave between #-oo# and #+oo#

graph{-x^2+4x+2 [-16.02, 16.02, -8.01, 8.01]}