(1) The product is zero when any of the factors are zero.
So 1-2ˣ=0 when 2ˣ=1, and x=0.
When -x²+2x+3=-(x²-2x-3)=-(x-3)(x+1)=0, x=3 or -1.
So we have three possible (critical) values for x: -1, 0, 3, therefore we consider the sign of f and g in 4 intervals: x<-1, x in (-1,0), x in (0,3) and x>3:
x<-1: f(x)<0, g(x)>0, f(x)g(x)<0;
x in (-1,0): f(x)>0, g(x)>0, f(x)g(x)>0;
x in (0,3): f(x)>0, g(x)<0, f(x)g(x)<0;
x>3: f(x)<0, g(x)<0, f(x)g(x)>0.
So f(x)g(x)≥0 when x=-1, x in (-1,0), x=0, x=3, x>3.
This can be summarised: -1≤x≤0, x≥3.
In words: x is between -1 and zero inclusive, or x is greater than or equal to 3.
(2) h(x)=f(x)+c=-x²+2x+3+c. The x-axis is h(x)=0, so:
h(x)=-x²+2x+3+c is a parabola (inverted U shape).
The equation can be h(x)=-(x²-2x-3-c)=-(x²-2x+1-4-c),
Let y=h(x)=-(x-1)²+4+c, y-(4+c)=-(x-1)².
The vertex (maximum in this case) is at (1,4+c). The vertex must lie on the x-axis (y=0) so 4+c=0 and c=-4.
h(x)=-x²+2x-1 or -(x-1)².
Red=f(x)×g(x), blue=h(x).