#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#
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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]}
