I have a test and need some desperate help on functions

[nando94]

I got a test and I am having trouble on functions. On the quiz we had this problem on composite functions.

given f(x) = 6/(3-x) and g(x) = 5/3x find the following

a. f of g

b. the domain of f of g

c. the range of f of g


My attempts

a. For a I got 6/3 - 5/3x = 18x/9x - 5 which was the right answer. No problems here. However I am stuck on b and c.

b.For b I got (-infinity, 0) (0, 5/9) (5/9, 3) (3,infinity). My teacher marked the 3's as wrong
and I don't know why. What I did was found the domain of f and g sperately and then excluded them from f of g. So I am kind of confused.

c. Right answer was (-infinity,0) (0,2) (2,infinity) For the range I kind of forgot how to find the range! But I found the inverse of f of g and then used the domain of the inverse since that's the range which was (-infinity,2)(2,infinity). Then I realized that since these were rational functions it was the horizontal asymptote but I am not sure if that's the right reason.

I really need help with this!

 

[eumyang]

a. For a I got 6/3 - 5/3x = 18x/9x - 5 which was the right answer. No problems here. However I am stuck on b and c.

Oh, really? What you wrote is
$\frac{6}{3} - \frac{5}{3x} = \frac{18x}{9x} - 5$
... and that's not the answer. (In other words, PLEASE use parentheses!)

b.For b I got (-infinity, 0) (0, 5/9) (5/9, 3) (3,infinity). My teacher marked the 3's as wrong
and I don't know why. What I did was found the domain of f and g sperately and then excluded them from f of g. So I am kind of confused.

The domain of g (the inner function) is (-∞, 0) U (0, ∞). The range is the same. The domain of f (the outer function), however, is (-∞, 3) U (3, ∞). The domain of f(g(x)) consists of all x-values in the domain of g that map to g(x)-values in the domain of f.

The problem here is that the range of g includes one value, 3, that cannot be in the domain of f. So you need the x-value so that g(x) = 3, and remove it from the domain of g. So solving
$\frac{5}{3x}=3$
gives you x = 5/9, and so the domain of f(g(x)) is
(-∞, 0) U (0, 5/9) U (5/9, ∞).

c. Right answer was (-infinity,0) (0,2) (2,infinity) For the range I kind of forgot how to find the range! But I found the inverse of f of g and then used the domain of the inverse since that's the range which was (-infinity,2)(2,infinity). Then I realized that since these were rational functions it was the horizontal asymptote but I am not sure if that's the right reason.

Now that we removed x = 5/9 from the domain of g, its range now becomes
(-∞, 0) U (0, 3) U (3, ∞). But since the domain of f was (-∞, 3) U (3, ∞) (which included 0), we have to now find what f(0) is and remove it from the range of f.
$f(0) = \frac{6}{3-0} = 2$
Since the range of f was (-∞, 0) U (0, ∞), the range of f(g(x)) now becomes
(-∞, 0) U (0, 2) U (2, ∞).

If you think this was all confusing, you'd be right. I've just taught this to my precalculus students and most of them had trouble understanding it all.

 

[nando94]

Thanks I see it now. You don't include the 3 because your finding f(g(x)). So you have to find g(3) instead. So if if you had to do g(f(x)) then you would find f(0) instead right?

 

[eumyang]

Thanks I see it now. You don't include the 3 because your finding f(g(x)). So you have to find g(3) instead.

Not quite. I said that you have to find x-values such that g(x) = 3. It's not the same thing as finding g(3).

So if if you had to do g(f(x)) then you would find f(0) instead right?

Again, not quite. Finding the domain of g(f(x)) is easier in this problem.
The range of f (now the inner function) is (-∞, 0) U (0, ∞).
The domain of g (now the outer function) is also (-∞, 0) U (0, ∞).
All values in the range of f are in the domain of g, so we don't have to "pull values out" of the domain of f.
So the domain of g(f(x)), in this case, is the same as the domain of f(x): (-∞, 3) U (3, ∞).
And since nothing was changed in the domain of g(f(x)), the range of g(f(x)) is the same as the range of g(x): (-∞, 0) U (0, ∞).