I have never quite understood how to evaluate domains and ranges of functions. I specifically have trouble with ranges.
For simple ones, like quadratic equations, I can usually find the minimum point and give a correct answer.
But take, for example:$$ f(x) = \frac{3}{2x+1}, x > 0$$and$$g(x) = \frac{1}{x} + 2, x > 0$$
I am so confused with the whole process of finding the ranges of functions, including those above as samples, that I can't even quite explain what or why.
Can someone please, in a very step-by-step process, detail exactly what steps you would take to get the ranges of the above functions? I tried substituting x-values (such as 0), and came up with $f(x) > 3$, but that was mostly guesswork -- also, $f(x) > 3$ is incorrect.
Also, is there an outline I can follow -- even for the thinking process, like check if A, check if B -- that would work every time?
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$\begingroup$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 assumethat 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 requiressignificant 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 sowithout 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 includeevery 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 notgraph the function without picking up your pencil. Therefore $s(x)$ isnot 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) itseems 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
$$ 2 < g(x). ~\text{Notice that} ~2 ~\text{is not part of the range.}$$
$\endgroup$ 4 $\begingroup$I added an additional explanation for the limits part at the end.
For $f$ you have that $x>0\Rightarrow 2x>0\Rightarrow2x+1>1\Rightarrow\dfrac{1}{2x+1}<1\Rightarrow\dfrac{3}{2x+1}<3$. On the other hand, $2x+1>1>0\Rightarrow\dfrac{1}{2x+1}>0\Rightarrow\dfrac{3}{2x+1}>0$. So $0<f(x)<3$ for every $x>0$. Furthermore, $\lim_{x\to0^+}f(x)=3$ and $\lim_{x\to+\infty}f(x)=0$, so, since $f$ is continuous in $(0,+\infty)$, we have that the range of $f$ is all of $(0,3)$.
For $g$ is similar. We have that for $x>0\Rightarrow\dfrac{1}{x}>0\Rightarrow\dfrac{1}{x}+2>2$. So $g(x)>2$ for every $x>0$. Furthermore, $\lim_{x\to0^+}g(x)=+\infty$ and $\lim_{x\to+\infty}g(x)=2$, so, since $g$ is continuous in $(0,+\infty)$, we have that the range of $g$ is all of $(2,+\infty)$.
I've been told that maybe I shouldn't use limits to justify the ranges, so I'll explain a little bit what I would say with words in the last part of the first paragraph if I didn't use limits.
For $f(x)=\dfrac{3}{2x+1}$ we have that $0<f(x)<3$ for every $x>0$. Now, we can see that, as $x$ goes closer and closer to $0$ then the $2x$ part of the denominator starts to get smaller and smaller, almost vanishing. If we could take $x=0$ then we would get $3$ as a value, but since we can't we can just say that the values of $f$ keep getting closer and closer to $3$; thus the parenthesis for the $3$ in the range.
If we make $x$ go bigger and bigger then $2x+1$ gets bigger and bigger, making the fraction $\dfrac{3}{2x+1}$ go smaller and smaller, almost vanishing into $0$. It never actually becomes $0$, but it keeps getting closer from above (from the positive numbers); thus the parenthesis for $0$ in the range. $f$ is a good function in $(0,+\infty)$: there are no "explosions" in the denominator, and it's defined always by the same expression.
This is not sufficient to say that $f$ is continuous, but it does look good. If you have the tools or the information so you can assert $f$ is indeed continuous, use it, since then $f$ can't jump from one value to another one, it has to go continuously from one to the other, taking all values in between. This is the same as thinking that if you want to get to a certain height from the ground, you'll have to be at some moment at ever height in between. Then $f$ takes every value between $0$ and $3$, since it can be as close to those values as we want.
Now you can try this reasoning with $g$.
$\endgroup$ 11 $\begingroup$Let's think about the example you gave: f(x) = 3/(2x + 1), with domain x > 0
To answer your first question, there are two "domains" of a function: (1) a given set of inputs to consider, and (2) every possible input that doesn't blow up in your face. In the case of this f(x), we say "for x > 0", and so this is the first case. We could plug in any value of x so long as 2x + 1 is not equal to zero.
Since we are strictly considering x > 0, we can manipulate this inequality to infer a similar inequality on f(x): namely, f(x) < 3. This is done by the manipulations:
- x > 0,
- 2x > 0,
- 2x + 1 > 1
- 1 > 1/(2x + 1) (careful with whether > or < here)
- 3 > 3/(2x + 1), so 3 > f(x).
This is certainly true, but it isn't the entire story. If consider a point in our range y = f(x) for some x > 0, then what is the corresponding value for x? This is the inverse function. We can certainly solve for it:
- 3/(2x + 1) = y
- 3/y = 2x + 1
- 3/y - 1 = 2x
- x = 3/(2y) - 1/2.
For example, if I know that f(x) = 1, then the value of x must be 3/2 - 1/2 = 2/2 = 1. Now consider: what value of y am I not allowed to plug in? Since dividing by zero is not allowed, we cannot have y = 0, meaning that there is no x such that f(x) = 0.
Now, if y is not zero, then it must be either (1) y > 0 or (2) y < 0. However, if y < 0, then 3/(2x+1) < 0. Hence, 3/(2x+1) < 0, and multiplying by (2x + 1), we get a false statement: 3 < 0 (again, careful with whether > or < here) So (2) cannot be true, and (1) must be true: y > 0, that is, f(x) > 0.
Combining these results: 3 > f(x) > 0. This illustrates my process for finding the range of a function. First, can you make any inferences about where f(x) could be using your domain? I showed that if x > 0 then 3 > f(x). Second, if you can find it, use the inverse function to try and pin down where f(x) lives.
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