What is the range of the composite function gf(A)=C?

[thereddevils]

Homework Statement

The sets A and B are defined respectively by

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

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

and the functions f and g are defined respectively by

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

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

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

Homework Equations

The Attempt at a Solution

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

so the range is (1, infinity)

 

[Cyosis]

That doesn't seem right. Are you sure you have defined your sets A and B correctly? As it is now the domain of B includes 1, but 1 is not a valid value, because g(1) is undefined.

Secondly if you plug a function f that sends A to B into another function g then g should take values from B as input. You on the other hand have g taking values from A as input.

Lastly if the domain of g o f is [0,1) then the range you found is certainly wrong. g o f is a function that increases in the interval [0,1) so the minimum value of the range cannot be 1 since (g o f)(0)>1.

 

[thereddevils]

That doesn't seem right. Are you sure you have defined your sets A and B correctly? As it is now the domain of B includes 1, but 1 is not a valid value, because g(1) is undefined.

Secondly if you plug a function f that sends A to B into another function g then g should take values from B as input. You on the other hand have g taking values from A as input.

Lastly if the domain of g o f is [0,1) then the range you found is certainly wrong. g o f is a function that increases in the interval [0,1) so the minimum value of the range cannot be 1 since (g o f)(0)>1.

ok , i see my mistake , so the correct range should be [4 , infinity) ? But the answer given is 2/3<=x<=4/3

 

[Cyosis]

You didn't answer all my questions:

That doesn't seem right. Are you sure you have defined your sets A and B correctly? As it is now the domain of B includes 1, but 1 is not a valid value, because g(1) is undefined.

Secondly if 2/3 is in the range of g o f then 2/3=1+3/(x-1)^2 has a solution for x which is in the domain of g o f. Check that this is not the case.