If g(x)=x+3 h(x)=9-x^2, what is the maximum of h(g(x))?

Answer 1

h = 9 - (x+3)^2

= 9 - x^2 -6 x - 9

= -x^2 - 6x

slope = dh/dx = -2x -6
zero when x = -3

curvature = d^2h/dx^2 = -2 so that is a maximum
then what is h?

-x^2 - 6 x = -x (x+6)
when x = -3
3 (3) = 9

which we could have said in the beginning because x^2 is always +
so
9-x^2 is always 9 or less :)