It is tough to come up with a univeral checklist, but I will
take a stab at it.
The first question is, is the function [e.g. $f(x)$]
well defined for each value in the domain. For example:
With respect to $f(x)$, if the constraint had allowed $x = -\frac{1}{2}$
then you would have a problem. Since the domain is restricted
to $x > 0$, for any finite value of $x$, the function is well defined.
With respect to $g(x)$, you have a virtually identical consideration.
Since the domain does not allow $x=0, g(x)$ is well defined throughout
its domain.
The next question is whether, in your opinion, the function is continuous.
The reason that this is important, is that if you are examining (for example)
some function $h(x)$, and you know that $h(x)$ is continuous, and you also
know that you can identify two points $x_1, x_2$ in the domain of $h(x)$,
then you know that the range must contain every value between
$h(x_1)$ and $h(x_2).$
The next step is to identify the endpoints of the domain, and consider
whether each endpoint is or is not in the domain.
With both of your functions, the lower endpoint is $0$, and the endpoint
is not in the domain. Similarly, the upper endpoint of both functions
is $\infty$, which is not a number, but rather a symbol that the upper
end of the domain is unbounded.
The way that I would handle the endpoints, is to pretend that you
have two positive real numbers $a,b$ with $a < b$, and you are first
considering the domain as all $x$ such that $a \leq x \leq b$.
Then, you consider the actual domain as the limit of the domain
determined in the previous paragraph, as $a \to 0$ and $b \to \infty$.
Since you are after a checklist, rather than the actual answers with
respect to the specific functions $f$ and $g$, that you provided, this
should be sufficient for you to solve this particular problem and also
work with similar problems in the future.
Please leave a comment directly below my answer, if you have any questions.
I will automatically be flagged.
Addendum
Response to comments/questions of : A Level Student : re 10-10-2020
First of all, I'd like to excerpt a point that amwhy commented to Darsen's answer:
Typically questions in algebra-precalculus to not presume knowledge of limits.
And my response:
@amWhy +1 (also) on your comment, which I agree with. However, it is hard to
fathom how the OP is supposed to systematically evaluate the range of values
given the domain of $0 < x$, without consideration of limits, as per my answer.
What I mean by this, is it is hard for me to conjure a systematic method,
as per your request, without assuming (in this case wrongly) that you
have a working knowledge of limits. What I will do, is continue to assume
that you understand limits, at least within the confines of my answer.
However I invite further questions. Please continue to post them directly
after my answer.
What I am going to do is respond to the specific questions that you raised in
your comments first. Then, I will pretend that I was assigned the problem
and describe how I would identify the range of $f(x)$ and the range of $g(x)$.
- So I tried to apply the method. We established that f(x) is continuous
(and well-defined, because it gives no exceptions, right?). Then we established
that the lower endpoint of my DOMAIN is 0, not included, and the upper endpoint
is infinity. After that, we pick two numbers (a<b) within our defined domain.
For this portion of your first comment, I agree with everything so far, except that it is not
a good idea to think of the upper endpoint as equaling infinity. This comes
too close to regarding infinity as a number, rather than a symbol. You would
do better to say that the upper endpoint is unbounded.
So I pick a=4 and b=8. f(4)=1/3, and f(8)=3/17. ...
Your confusion here is my fault for not providing clear details about what
to do with the idea that you first pretend that the domain is $a \leq x \leq b$,
and then examine what happens as $a \to 0$, and $b \to \infty.$
See the later portion of my addendum, where I provide better details about
specifically attacking the problem you posted in your query.
...... so that 0 is my next endpoint? So I write it out like this: ...
Again, see the later portion of my addendum.
What's the difference between well-defined and continuous?
The best way to answer this question is to first examine this question with
respect to the sample functions in your original query. Then, make more general
comments. Your functions are:
$$ f(x) = \frac{3}{2x+1}, x > 0$$
and
$$g(x) = \frac{1}{x} + 2, x > 0$$
Consider what would happen if you tried to examine $f(x)$ for any real number
$x$, rather than only those values of $x$ that are greater than $0$. $f(x)$ has
$(2x+1)$ in its denominator, and division by $0$ is forbidden. Therefore,
$f(x)$ is not well defined at $x = \frac{-1}{2}.$
Similarly, if you tried to examine $g(x)$ for any real number $x$, you would
find that $g(x)$ is not well defined at $x=0$.
So the concept of well defined means that a function [e.g. $f(x)$] is not well defined at
a specific value $x = x_0$ if you are unable to assign a specific (finite)
value to $f(x_0)$.
Continuity is an entirely different subject. Understanding this subject requires
significant experience with limits. The best that I can do at this point is to
give you an intuitive, informal idea of what Continuity is supposed to represent.
If you have a function with a specific domain (for example all $x$ such that $x > 0$),
and you are wondering whether the function is continuous throughout the domain,
simply ask yourself this: when I go to manually graph the function, can I do so
without picking up my pencil.
With both of the functions that you provided, $f(x)$ and $g(x)$, the answer is
yes. Therefore, you are able to assume that (for example), that if $x_1$ and $x_2$
are in the domain of $f(x)$, with $x_1 < x_2$, then the range of $f(x)$ must include
every value between $f(x_1)$ and $f(x_2)$.
Informally, this is because as you are graphing the function with your pencil,
as your pencil travels from $f(x_1)$ to $f(x_2)$, you are not allowed to
pick up your pencil. Therefore the graph must traverse every point between
$f(x_1)$ and $f(x_2)$.
This begs the question: how in the world do you try to determine whether a specific
function is or is not continuous? This is an enormously complicated question,
well beyond what I can address in my answer. However, I can give you an intuitive
idea of what a discontinuous function might look like.
Consider the function $s(x)$ defined on the domain $0 < x < 3$ as follows:
$$\text{For} ~0 < x < 1, s(x) = 0.$$
$$\text{For} ~1 \leq x < 2, s(x) = 1.$$
$$\text{For} ~2 \leq x < 3, s(x) = 2.$$
A reasonable way of thinking about this function $s(x)$ is that you can not
graph the function without picking up your pencil. Therefore $s(x)$ is
not continuous throughout its domain of $0 < x < 3$.
Also, if my function is not well-defined, or continuous ...
I can't use the above method? But I can still find the range?
Yes absolutely. But providing a systematic way of doing so, in these
more complicated situations is well beyond what I can cover in this answer.
So for the purposes of this answer, it is important to
Verify that $f(x)$ and $g(x)$ are each well defined in the domain of $0 < x$,
which they are.
Presume, that $f(x)$ and $g(x)$ are continuous, because (very informally) it
seems that you can graph each of these functions without picking up your pencil.
After accepting the presumption of continuity, realizing that as a consequence,
that (for example) if $x_1$ and $x_2$ are both in the range of $f(x)$, with $x_1 < x_2$, then the range of $f(x)$ must include
all values between $f(x_1)$ and $f(x_2)$.
$\underline{\text{attacking the problem you posted in your query}}$
I will attack them one at a time.
$$ f(x) = \frac{3}{2x+1}, x > 0$$
Start with the pretense that the domain of $x$ is $a \leq x \leq b$,
with $a < b$, where for the moment, $a$ and $b$ are going to be regarded
as unspecified values (i.e. variables).
Then,
$$ f(a) = \frac{3}{2a+1} ~\text{and} ~f(b) = \frac{3}{2b+1}$$
Furthermore, with $f(x) = \frac{3}{2x+1}$, ask yourself:
As $x$ increases, is $f(x)$ strictly decreasing?
Answer:
Yes, because as $x$ increases, the denominator will increase, which
causes the value of $f(x)$ to decrease.
Therefore, the maximum value for $f(x)$ will be achieved as $x$ is allowed
to decrease as much as possible.
As $a$ approaches $0$ from above, $f(x)$ will aproach
$$f(x) = \frac{3}{2[0]+1} = \frac{3}{1} = 3.$$
Similarly, as $b$ increases in an unbounded manner
$f(x)$ will be decreasing. What is the smallest value that $f(x)$ can achieve?
As $b$ increases, the denominator will be able to take on any value. Therefore,
the denominator of $f(x)$ will be unbounded.
If you have a fraction $\frac{P}{Q}$, where $P > 0$ is a fixed value, and $Q$ can
get arbitrarily large, then the fraction will go to zero. That is, as $Q$
grows, any small positive value of $\frac{P}{Q}$ can be achieved.
Using this concept with respect to $f(x)$, we see that as $b$ grows unbounded,
$f(x)$ will approach $0$ from above. This means that any small postive
value will fall within the range of $x$.
Therefore, in conclusion, for $f(x)$ the range of $f(x)$ is:
$$0 < f(x) < 3.$$
Notice that I used the phrase "$b$ grows unbounded", rather than the terminology
$b \to \infty$. Although the terminology is appropriate, I deliberately avoided
using it simply to emphasize the idea that infinity is a symbol rather than
a number. Infinity symbolizes unbounded growth.
At this point, you have a valid claim of Foul.
You asked for a systematic method. The approach that I used, after pretending
that the domain was $a \leq x \leq b$ and examining what happens when
$a \to 0$ and $b \to \infty$ was anything but systematic.
I regard this as a very valid complaint. Unfortunately, given the problems
that you will be assigned in the future, I am unable to provide a rigorous
checklist of the steps that you take. These steps will depend on the problems
that you are assigned.
The best that I can do is provide the worked examples for the specific $f(x)$
and $g(x)$ that you gave. The idea is for you to use these worked examples
as a guide; this is the best that I can do.
To finish, I will similarly attack $g(x)$ :
$$g(x) = \frac{1}{x} + 2, x > 0$$
I am going to use virtually the same ideas that I used when attacking $f(x)$.
If the domain were $a \leq x \leq b$ then,
$$g(a) = \frac{1}{a} + 2 ~\text{and} ~g(b) = \frac{1}{b} + 2.$$
Again, in $g(x)$, $x$ is occuring in the denominator only, so
as $x$ increases, $g(x)$ decreases and as $x$ decreases, $g(x)$ increases.
As $a \to 0, ~\frac{1}{a}$ will $\to \infty$ (i.e. grow unbounded). Therefore,
the upper end of the range of $g(x)$ is unbounded.
As $b \to \infty, \frac{1}{b}$ will go to $0$.
Therefore, as $b \to \infty,$ $g(b)$ will go to $0 + 2$.
It is critical to note that for any finite value of $b, g(b) > 2.$
Therefore, the range of $g(x)$ is
$$ 2 < g(x). ~\text{Notice that} ~2 ~\text{is not part of the range.}$$
Best Answer
The problem comes when you take the reciprocal. The step $2-x^2 \leq 2$ is correct. When you invert, however, you have to look at 2 cases; when $2-x^2 \geq 0$ and $2-x^2 \leq 0$. When you take the reciprocal, you assume that the first case is correct. The second case, however, gives rise to the other part of the range.