# Verify if f(g(x))>=0 for f(x)=x^2+2x+1; g(x)=x-2011

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f(x)= x^2 + 2x + 1

g(x) = x- 2011

We need to verify if fog(x) is zero or positive.

Let us determine the function fog(x) = f(g(x))

==> f(g(x))= f(x-2011) = (x-2011)^2 + 2(x-2011)+1

==> f(g(x)) = x^2 - 4022x + 2011^2 + 2x - 4022 + 1

==> f(g(x))= x^2 - 4020x + 4040100

==> f(g(x))= (x-2010)^2

We notice that (x-2010)^2 is always positive or zero when x=2010

Now we proved that f(g(x) > 0 for all R-{ 2010}

Also f(g(x))= = 0 for x = 2010.

**Then f(g(x)) >= 0 **

We'll recall the rule of composition of 2 functions:

f(g(x)) = (g(x))^2 + 2*g(x) + 1

We'll recognize a perfect square in the expression of f(x):

f(x) = (x+1)^2

According to this, we'll have:

f(g(x)) = [g(x) + 1]^2

**Since a binomial raised to square is always positive, the inequality f(g(x)) >= 0 is verified.**