How to find the range of a function with square roots?

[arham_jain_hsr]

Homework Statement

Find the domain and range of f(x)=√(9-x^2)

Relevant Equations

N/A

$$y = f(x) = \sqrt{9-x^2}$$
According to me,
Domain: $$ 9-x^2 \geq 0 \implies (x+3)(x-3) \leq 0 \implies x \in [-3,3] $$
which is correct, but this is how I calculate the range:
$$y = \sqrt{9-x^2} \implies y^2 = 9-x^2 \implies x^2 = 9-y^2$$
Now, since $$ 9-x^2 \geq 0 $$
we get $$9-9+y^2 \geq 0 \implies y^2 \geq 0 \implies y \geq 0$$
$$\therefore y\in[0,\infty)$$
But, the correct answer is supposed to be [0, 3]. Where did I go wrong in my approach? Also, how do I tackle similar problems algebraically?
The reason I think the answer "should" have been correct is that I think that y is only restricted by values of x. So, if we use the domain as a constraint on y, we should get the range of y. But, somehow that yields the wrong answer...

 

[PeroK]

In general, an implication is not reversible. If you conclude that ##\Rightarrow \ f(x) \ge 0##, then that does not mean that the range of ##f## is ##[0, \infty)##. For example, if ##f(x) = 1##, then ##f(x) \ge 0##, but the range of ##f## is only the single point ##1##.

The range is precisely the set of possible functions values. Not any set which contains those function values.

PS to emphasise the point, for any function you can conclude that ##\Rightarrow \ f(x) \in \mathbb R##.

 

[PeroK]

PS if the range of ##f## is ##[0, \infty)##, the you should be able to find ##x## such that ##f(x) = 4##. Is that possible?

 

[arham_jain_hsr]

In general, an implication is not reversible. If you conclude that ##\Rightarrow \ f(x) \ge 0##, then that does not mean that the range of ##f## is ##[0, \infty)##. For example, if ##f(x) = 1##, then ##f(x) \ge 0##, but the range of ##f## is only the single point ##1##.

The range is precisely the set of possible functions values. Not any set which contains those function values.

PS to emphasise the point, for any function you can conclude that ##\Rightarrow \ f(x) \in \mathbb R##.

My idea was that there is only one condition on what y is like. And, therefore after applying that condition, whatever interval we get for y should be the required range.

Like, in my case, I thought the range of y is [itex][0, \infty)[/itex] because I didn't know anything else about y (other than its relation to x) that could possibly narrow the range down.

PS if the range of ##f## is ##[0, \infty)##, the you should be able to find ##x## such that ##f(x) = 4##. Is that possible?

I supposed that by "using" [itex]9 - x^2 > 0[/itex] to find my answer, the interval of y that I get would be such that it complied with [itex]9 - x^2 > 0[/itex].

 

[PeroK]

My idea was that there is only one condition on what y is like. And, therefore after applying that condition, whatever interval we get for y should be the required range.

Like, in my case, I thought the range of y is [itex][0, \infty)[/itex] because I didn't know anything else about y (other than its relation to x) that could possibly narrow the range down.

You know a lot more about ##y## than it's just any non-negative number.

I supposed that by "using" [itex]9 - x^2 > 0[/itex] to find my answer, the interval of y that I get would be such that it complied with [itex]9 - x^2 > 0[/itex].

Exactly! Or, sketch a graph of the function.

 

[MatinSAR]

Homework Statement: Find the domain and range of f(x)=√(9-x^2)
Relevant Equations: N/A

Also, how do I tackle similar problems algebraically?

You can also start from ##-x^2 \leq 0 ## then try to create ##f(x)## on one side of this inequality the other side will help you find the range.

 

[arham_jain_hsr]

You know a lot more about ##y## than it's just any non-negative number.

Exactly! Or, sketch a graph of the function.

But, why does my approach yield the wrong answer?

 

[WWGD]

Maybe if you set ##y=f(x)##, square both sides, you'll obtain a revognizable expression that will help you determine the range.

 

[PeroK]

But, why does my approach yield the wrong answer?

I already explained that. The range is the precise set of values that ##f(x)## can take. You didn't look for the precise set of values. You just noted that the range is a subset of ##[0, \infty)## and then stopped there. You didn't use the other information about ##f##.

 

[PeroK]

PS to take an example. Find the range of the function ##f(x) = x^2 + 1##?

Your solution: ##f(x) > 0##, so the range is ##[0, \infty)##.

My solution: the range is precisely ##[1, \infty)##.

 

[WWGD]

@arham_jain_hsr : You may want to read up on differences between range and codomain.
You were so close in your initial post. Your ##x,y## values are connected to each other by the expression

## x^2+ y^2=9##

Do you recognize the collection of points ##(x,y)## that satisfy the above relation?
https://en.m.wikipedia.org/wiki/Codomain
https://math.stackexchange.com/ques...difference-between-codomain-and-range#3317971

 

[arham_jain_hsr]

@arham_jain_hsr : You may want to read up on differences between range and codomain.
You were so close in your initial post. Your ##x,y## values are connected to each other by the expression

## x^2+ y^2=9##

Do you recognize the collection of points ##(x,y)## that satisfy the above relation?
https://en.m.wikipedia.org/wiki/Codomain
https://math.stackexchange.com/ques...difference-between-codomain-and-range#3317971

A circle at origin with radius 3 units?

 

[WWGD]

A circle at origin with radius 3 units?

Precisely.

 

[arham_jain_hsr]

Precisely.

Oh, so, since it represents a circle, y may vary from -3 to +3. We are getting this because we lost the information that y must be positive in the process of squaring both sides. So, reconsidering that, we get the range [0, 3].

 

[arham_jain_hsr]

Well, I think my confusion can be better stated as follows:
Why, if I derived my answer from the fact that "[itex]9 - x ^ 2[/itex] should be greater than equal to [itex]0[/itex]", am I getting an answer that allows values of [itex]y[/itex] for which [itex]x[/itex] is greater than [itex]3[/itex]?

 

[WWGD]

Well, I think my confusion can be better stated as follows:
Why, if I derived my answer from the fact that "[itex]9 - x ^ 2[/itex] should be greater than equal to [itex]0[/itex]", am I getting an answer that allows values of [itex]y[/itex] for which [itex]x[/itex] is greater than [itex]3[/itex]?

Well, ##9-x^2 \geq 0 ## doesn't exclude values for ##x## in ##[-3,0]##.

 

[DaveE]

I think it would be helpful for you to review inequality equations. There are some good tutorials at this site:
https://www.khanacademy.org/math/cc...-two-step-inequalities/v/solving-inequalities

 

[PeroK]

Well, I think my confusion can be better stated as follows:
Why, if I derived my answer from the fact that "[itex]9 - x ^ 2[/itex] should be greater than equal to [itex]0[/itex]", am I getting an answer that allows values of [itex]y[/itex] for which [itex]x[/itex] is greater than [itex]3[/itex]?

I already explained that. You simply showed that ##y \ge 0##. That's true, but not the whole story. I'll repost post #2, which you either didn't read or didn't understand.

In general, an implication is not reversible. If you conclude that ##\Rightarrow \ f(x) \ge 0##, then that does not mean that the range of ##f## is ##[0, \infty)##. For example, if ##f(x) = 1##, then ##f(x) \ge 0##, but the range of ##f## is only the single point ##1##.

The range is precisely the set of possible functions values. Not any set which contains those function values.

PS to emphasise the point, for any function you can conclude that ##\Rightarrow \ f(x) \in \mathbb R##.

 

[arham_jain_hsr]

Oh, ok. Now, I understand. As it appears, my flaw here is pretty silly. So, basically, to find the domain, I "used" the fact that [itex]y^2 = 9 - x^2 \geq 0[/itex]. That's fine, because the fact that [itex]y^2[/itex] should be greater than [itex]0[/itex] puts a constraint on what the possible values of [itex]x[/itex] can be.
Now, here, I am trying to use the same piece of information [itex]y^2>0[/itex] (which due to the substitution, I was illusioned to think is something "novel") to find the range. But, the problem is that this piece of information doesn't restrict the values of [itex]y[/itex] to what is supported by the possible values of [itex]x[/itex].
And, hence unless and until I "constrain" the values of [itex]y[/itex] as described by the question, I won't get the correct range. For example, if I'd rather begun from [itex]x^2 = 9 - y^2 > 0[/itex], therefore restricting the values of [itex]y[/itex] "based" on the possible values of [itex]x[/itex], I'd indeed have arrived at the correct range for this relation. (Of course, limiting the values of y to positive real numbers only)

Thank you so much for helping me understand @PeroK @MatinSAR @WWGD @DaveE :)

 

[Mark44]

As it appears, my flaw here is pretty silly. So, basically, to find the domain, I "used" the fact that [itex]y^2 = 9 - x^2 \geq 0[/itex]. That's fine, because the fact that [itex]y^2[/itex] should be greater than [itex]0[/itex] puts a constraint on what the possible values of [itex]x[/itex] can be.

I'm not sure you understand, yet. ##y^2##, merely by virtue of being the square of a real number, will always be greater than or equal to zero. The same is true of ##x^2##. This means that the largest value of ##y^2## is 9, hence ##y \in [-3, 3]##. But since the original equation is ##y = \sqrt{9 - x^2}##, that means the smallest value of y is 0.

Now, here, I am trying to use the same piece of information [itex]y^2>0[/itex] (which due to the substitution, I was illusioned to think is something "novel") to find the range. But, the problem is that this piece of information doesn't restrict the values of [itex]y[/itex] to what is supported by the possible values of [itex]x[/itex].

 

[PeroK]

I agree with @Mark44. I would have noted that:
$$0 \le \sqrt{9-x^2} \le 3$$Informally, you can then conclude that the range of ##f## is ##[0,3]##. Technically, however, you still have to show that for any ##y \in [0,3]## you can find ##x## such that ##f(x)=y##.

 

[arham_jain_hsr]

I'm not sure you understand, yet. ##y^2##, merely by virtue of being the square of a real number, will always be greater than or equal to zero. The same is true of ##x^2##. This means that the largest value of ##y^2## is 9, hence ##y \in [-3, 3]##. But since the original equation is ##y = \sqrt{9 - x^2}##, that means the smallest value of y is 0.

I'm sorry. I don't get you. Isn't that exactly what I said as well?

 

[Mark44]

I'm sorry. I don't get you. Isn't that exactly what I said as well?

It's hard for me to say, but I don't think so. Here's what you said:

Now, here, I am trying to use the same piece of information
[itex]y^2>0[/itex] (which due to the substitution, I was illusioned to think is something "novel") to find the range. But, the problem is that this piece of information doesn't restrict the values of [itex]y[/itex] to what is supported by the possible values of [itex]x[/itex].
And, hence unless and until I "constrain" the values of [itex]y[/itex] as described by the question, I won't get the correct range. For example, if I'd rather begun from [itex]x^2 = 9 - y^2 > 0[/itex], therefore restricting the values of [itex]y[/itex] "based" on the possible values of [itex]x[/itex], I'd indeed have arrived at the correct range for this relation. (Of course, limiting the values of y to positive real numbers only)

You wrote ##y^2 > 0## which really should be ##y^2 \ge 0##, and this has nothing to do with any substitution, but only due to the fact that the square of a real number has to be nonnegative. Then you made a diversion to another equation, ##x^2 = 9 - y^2##, and seemed to miss the fact that ## 0 \le y = \sqrt{9 - x^2} \le 3##, which gives you the range.
As far as the domain, ##9 - x^2 \ge 0 \Rightarrow -3 \le x \le 3##.

 

[arham_jain_hsr]

It's hard for me to say, but I don't think so. Here's what you said:
You wrote ##y^2 > 0## which really should be ##y^2 \ge 0##, and this has nothing to do with any substitution, but only due to the fact that the square of a real number has to be nonnegative. Then you made a diversion to another equation, ##x^2 = 9 - y^2##, and seemed to miss the fact that ## 0 \le y = \sqrt{9 - x^2} \le 3##, which gives you the range.
As far as the domain, ##9 - x^2 \ge 0 \Rightarrow -3 \le x \le 3##.

Noted. Thanks for pointing that out.