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March 27, 2015

March 27, 2015

Posted by **Sarah** on Saturday, September 25, 2010 at 10:30am.

I was able to determine that:

f(g(x))= sin(x^2-x+1) and

g(f(x))= sin(x)^2 -sin(x) +1

Here is what I am thinking:

the range of f(x) is (x e R) and the range of g(x) is {x e R). Therefore the domain of f(g(x)) and g(f(x)) is {x e R)

the range of f(x) is {-1=< y=< 1} and the range of g(x) is {y>= .75}. Therefore the range of g(x) and f(x) will be what satisfies both the functions, { .75=< y =< 1}

Is this correct? If not, how would I go about solving this problem, specifically the range? Thanks for your help.

- Math -
**MathMate**, Saturday, September 25, 2010 at 6:24pmf(g(x))= sin(x^2-x+1) and

g(f(x))= sin(x)^2 -sin(x) +1

are correct.

The range of f(x)=sin(x) is [-1,1].

The range of g(x)=x²-x+1 is indeed ℝ.

However, the domain of both functions is ℝ.

So the domain of f(g(x)) ℝ and the range is still [-1,1].

The domain of g(f(x)) is ℝ. However, since sin(x) is a periodic function, we just have to limit the search to [-2π,2π].

Find the absolute minimum and absolute maximum on the interval [-2π,2π] and that would be the range (all possible values of the function).

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