# Thread: range of a composite function

1. ## range of a composite function

The sets A and B are defined respectively by

$\displaystyle A={x\in R : 0\leq x\leq 1}$

$\displaystyle B={x\in R : 1\leq x\leq 2}$

and the functions f and g are defined respectively by

$\displaystyle f(x)=x^2-2x+2$

$\displaystyle g(x)=\frac{x+2}{x-1}$

where f(A)=B , g(B)=C with C as the range of the function g .

Find the range of the composite function gf(A)=C

Attempt :

The way they put it kinda confuse me .

$\displaystyle gf(x)=1+\frac{3}{(x-1)^2}$ , with domain [0,1)

so the range is (1, infinity)

2. It is true that, as f(x) approaches 1, gof(x) approaches infinity. However g(x) is never equal to 1. As f(x) goes to 2, gof(x) goes to 4. The range of gof is $\displaystyle [4, \infty)$.

You could also have see that by writing gof(x) as $\displaystyle \frac{x^2- 2x+ 4}{(x- 1)^2}$. When x= 0, gof(0)= 4.

3. Originally Posted by HallsofIvy
It is true that, as f(x) approaches 1, gof(x) approaches infinity. However g(x) is never equal to 1. As f(x) goes to 2, gof(x) goes to 4. The range of gof is $\displaystyle [4, \infty)$.

You could also have see that by writing gof(x) as $\displaystyle \frac{x^2- 2x+ 4}{(x- 1)^2}$. When x= 0, gof(0)= 4.
But the answer given is $\displaystyle \frac{4}{3}\leq x\leq \frac{3}{2}$