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

AI Thread Summary
The discussion revolves around finding the range of the composite function gf(A)=C, where sets A and B are defined, and functions f and g are given. There is confusion regarding the definitions of sets A and B, particularly concerning the inclusion of the value 1, which makes g(1) undefined. The initial calculation of the range as (1, infinity) is challenged, leading to the conclusion that the correct range should be [4, infinity). However, there is a discrepancy with the provided answer of 2/3<=x<=4/3, prompting further investigation into the validity of this range. The conversation emphasizes the importance of correctly defining the domains and ranges of the functions involved.
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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)
 
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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.
 
Cyosis said:
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
 
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.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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