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 $[0, \infty)$ 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" $9 - x^2 > 0$ to find my answer, the interval of y that I get would be such that it complied with $9 - x^2 > 0$.

 

[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 $[0, \infty)$ 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" $9 - x^2 > 0$ to find my answer, the interval of y that I get would be such that it complied with $9 - x^2 > 0$.

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 "$9 - x ^ 2$ should be greater than equal to $0$", am I getting an answer that allows values of $y$ for which $x$ is greater than $3$?

 

[WWGD]

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

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 "$9 - x ^ 2$ should be greater than equal to $0$", am I getting an answer that allows values of $y$ for which $x$ is greater than $3$?

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 $y^2 = 9 - x^2 \geq 0$. That's fine, because the fact that $y^2$ should be greater than $0$ puts a constraint on what the possible values of $x$ can be.
Now, here, I am trying to use the same piece of information $y^2>0$ (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 $y$ to what is supported by the possible values of $x$.
And, hence unless and until I "constrain" the values of $y$ as described by the question, I won't get the correct range. For example, if I'd rather begun from $x^2 = 9 - y^2 > 0$, therefore restricting the values of $y$ "based" on the possible values of $x$, 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 $y^2 = 9 - x^2 \geq 0$. That's fine, because the fact that $y^2$ should be greater than $0$ puts a constraint on what the possible values of $x$ 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 $y^2>0$ (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 $y$ to what is supported by the possible values of $x$.

 

[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
$y^2>0$ (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 $y$ to what is supported by the possible values of $x$.
And, hence unless and until I "constrain" the values of $y$ as described by the question, I won't get the correct range. For example, if I'd rather begun from $x^2 = 9 - y^2 > 0$, therefore restricting the values of $y$ "based" on the possible values of $x$, 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.