f(x)=x^2+2x+2 is defined for all Real values of x
rArr Domain is RR
f(x)=x^2+2x+2 has a minimum value of f(x)=1 but no maximum value
rArr Range is [1,+oo), RR
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
How do we know that f(x) has a minimum value of 1?
Since f(x) is a quadratic with a coefficient of x which is greater than zero,
we know that it has a minimum.
We can find this minimum by taking the derivative of f(x), setting that to zero, solving for x, and then using that value to find f(x) at that value:
color(white)("XXX")f'(x)=2x+2=0
color(white)("XXX")rarr x=-1
color(white)("XXX")rarr f(x=-1)=(-1)^2+2 * (-1)+2=1
If it helps here is a graph of f(x)=x^2+2x+2 to help verify this:
graph{x^2+2x+2 [-3.61, 2.55, -0.512, 2.568]}
