# Range of a function

## Homework Statement

Find the range of the function $$f(x) = \sqrt{4-x^{2}}$$

## The Attempt at a Solution

$$f(x)=\sqrt{4-x^{2}}; -2\leqx\leq2$$
$$f(2)=\sqrt{4-(2)^2}$$
$$f(2)=\sqrt{4-4}$$
$$f(2)=\sqrt{0}$$
$$f(2)=0$$

$$f(0)=\sqrt{4-(0)^2}$$
$$f(0)=\sqrt{4}$$
$$f(0)=2$$

Range of $$f=[0,2]$$

I believe my answer is correct. What I am wondering is whether I'm showing my work properly. If I was to hand this in to a prof/TA, would they dock me marks, and where?

Mentallic
Homework Helper
Yes it could be answered better. This question is very simple so logic was easily used in this case, but lets try use a process that would help for more complicated questions.

First state that the range must be more than or equal to 0, since square root. So find the domain.
i.e.
$$\sqrt{4-x^2}\geq 0$$

Now either by simply using a rough sketch of its graph, testing points, or some other means (or maybe even for your class' level, such a simple request is not expected), show $$-2\leq x\leq 2$$

Now, take the function under the square root, and use calculus or some other means (in this case, the vertex of the parabola can be found)

e.g. $$y=4-x^2$$

$$\frac{dy}{dx}=-2x$$

Find the turning point ($$\frac{dy}{dx}=0$$), show it is a maximum by the second derivative or stating that since it lies in the real domain of f(x), then it must be maximum, and now you have the range. $$0\leq f(x)\leq 2$$

Mentallic it would be much easy if he wrote:

$$4-x^2 \geq 0$$

$$(2-x)(2+x) \geq 0$$

Now (2-x)≥0 && (2+x)≥0 OR (2-x)≤0 && (2+x)≤0

Last edited:
Mentallic it would be much easy if he wrote:

$$4-x^2 \geq 0$$

$$(2-x)(2+x) \geq 0$$

Now (2-x)>0 && (2+x)>0 OR (2-x)<0 && (2+x)<0

How did you get that part?

Two negative numbers give positive one right?

Just to mention, your range and Mentallic's range is not complete.

Actually, my range and Mentallic's range are complete. If you don't know how to find the range properly, please don't bother posting.

As a sidenote, you weren't even trying to solve for the range. There is a difference between domain and range.

Actually, my range and Mentallic's range are complete. If you don't know how to find the range properly, please don't bother posting.

As a sidenote, you weren't even trying to solve for the range. There is a difference between domain and range.
Whoa! Again I misread the problem. So sorry.

Since $$f(x)=\sqrt{4-x^2}$$ is a continuous function, it is possible using calculus to determine the minimum and maximum values of f on its domain.

The domain is fairly easily determined to be [-2, 2].

Unfortunately without calculus it would be difficult to determine whether it is the case that f wanders above or below its values at -2 and 2.

For example $g(x)=x^3-x \text{ for } -1 \leq x \leq 1$ has values f(-1) = 0 and f(1) = 0, but the function has a range of $$[-2\sqrt{3}/9, 2\sqrt{3}/9]$$.

--Elucidus

Mark44
Mentor
Since $$f(x)=\sqrt{4-x^2}$$ is a continuous function, it is possible using calculus to determine the minimum and maximum values of f on its domain.
Given that this was posted in the Precalculus forum, a solution using calculus is probably inappropriate. I don't know the OP's background, but it's likely that he or she has not studied calculus yet, so any help or hints in that direction are probably misplaced.
The domain is fairly easily determined to be [-2, 2].

Unfortunately without calculus it would be difficult to determine whether it is the case that f wanders above or below its values at -2 and 2.
For this problem, calculus is unnecessary, unlike the problem you posed in the following example. For this problem, you can look at the behavior of the related function y = 4 - x2 to shape of the graph (and thereby determine the range). Then, one can look at the graph of $$y = \sqrt{4 - x^2}$$ to see how taking the square root affects the domain and range. No calculus needed.
For example $g(x)=x^3-x \text{ for } -1 \leq x \leq 1$ has values f(-1) = 0 and f(1) = 0, but the function has a range of $$[-2\sqrt{3}/9, 2\sqrt{3}/9]$$.

--Elucidus
If you give an example, it's a good idea to give one that is similar enough to the original problem that the same sort of solution applies. That way you can convey the basic underlying ideas that can be applied to the original problem.

If you give an example, it's a good idea to give one that is similar enough to the original problem that the same sort of solution applies. That way you can convey the basic underlying ideas that can be applied to the original problem.

I guess the point I was hoping to make was not made well. My comment was motivated by trying to indicate that some functions' ranges require more advanced methods (like calculus) - I was giving an example of one of the more advanced problems.

I have frequently seen students try methods they've learned in earlier courses on problems for which that method was inappropriate not knowing the limitations of the technique. I often try to explain these limitations when presenting new methods - including an example for which it fails.

I agree the range for $y = \sqrt{4-x^2}$ can fortunately determined geometrically as you say. This method does not always work though.

--Elucidus

Mark44
Mentor
I guess the point I was hoping to make was not made well. My comment was motivated by trying to indicate that some functions' ranges require more advanced methods (like calculus) - I was giving an example of one of the more advanced problems.
Right. And my point was that for a precalculus problem, the first technique to be explored or explained ought not to invoke techniques that involve calculus, but instead should appeal to the student's graphical sense or basic knowledge of functions. IMO, the simplest approach that will do the job is the one to start with. After the student understands that technique, then you can elaborate on other approaches, but not until then.
I have frequently seen students try methods they've learned in earlier courses on problems for which that method was inappropriate not knowing the limitations of the technique. I often try to explain these limitations when presenting new methods - including an example for which it fails.

I agree the range for $y = \sqrt{4-x^2}$ can fortunately determined geometrically as you say. This method does not always work though.

--Elucidus

Why not find the domain of the function's "inverse" and from that determine the range of the original function? I know it's not one-to-one, but it can help.