SOLUTION: Suppose you have a function y = f(x) such that the domain of f(x) is 1 ≤ x ≤ 6 and the range of f(x) is −3 ≤ y ≤ 5. What is the domain of f(2(x &#872

Algebra ->  Functions -> SOLUTION: Suppose you have a function y = f(x) such that the domain of f(x) is 1 ≤ x ≤ 6 and the range of f(x) is −3 ≤ y ≤ 5. What is the domain of f(2(x &#872      Log On


   



Question 403290: Suppose you have a function y = f(x) such that the domain of f(x) is 1 ≤ x ≤ 6 and the range of f(x) is −3 ≤ y ≤ 5. What is the domain of f(2(x − 3))
Found 2 solutions by jim_thompson5910, robertb:
Answer by jim_thompson5910(35256) About Me  (Show Source):
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Since the domain is 1%3C=x%3C=6, this means that the set of possible input values is {1,2,3,4,5,6}. Say for the sake of argument that we're only dealing with integers (to make things easy for us)


Now consider that instead of plugging in 'x', you plug in x-3. So this means that instead of plugging in x=1, you'll plug x-3=1-3=-2 into the function.

So this means that the domain of f(x-3) is now


1-3%3C=x-3%3C=6-3


which becomes


-2%3C=x-3%3C=3


So the domain of f(x-3) is -2%3C=x-3%3C=3


Now we're going to take it a step further. Instead of plugging in x-3, we're going to double it and plug in 2(x-3). So instead of plugging in -2, we're going to plug (-2)(2)=-4 into the function.

So this means that the domain becomes


2%28-2%29%3C=2%28x-3%29%3C=2%283%29


-4%3C=2%28x-3%29%3C=6


So the domain of f(2(x-3)) is -4%3C=2%28x-3%29%3C=6 given that the domain of f(x) is 1%3C=x%3C=6.


Basically, the smallest number that the input 2(x-3) can be is -4 and the largest number that 2(x-3) can be is 6

Answer by robertb(5830) About Me  (Show Source):
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The domain of f(2(x-3))can be determined in all generality, not only using discrete values.
Let g(x) = 2(x-3) = 2x - 6.
Then f(2(x-3)) = (f o g)(x) = f(g(x)). Let the domain of g be D%5Bg%5D, its range R%5Bg%5D. Similarly let the domain of f be D%5Bf%5D, its range R%5Bf%5D.
Then to find the domain of f o g we must find all x values in D%5Bg%5D (which is the set of all real numbers), that are the pullback of R%5Bg%5DintersectionD%5Bf%5D. Since the range of g is represented by 2(x-3), and the domain of f is the closed interval [1,6], we must then have
1+%3C=+2%28x+-+3%29+%3C=+6
<==> 1%2F2+%3C=+x+%0D%0A-+3+%3C=+3
<==> 7%2F2+%3C=+x+%3C=+6.
Thus the domain of (f o g)(x) = f(2(x-3)) is the closed interval [7/2, 6].