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

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To find the range of the function y = f(x) = √(9 - x²), the correct approach involves recognizing that the domain is restricted to x ∈ [-3, 3], which leads to y being constrained to values between 0 and 3. The initial misunderstanding arose from incorrectly concluding that y could take any non-negative value, leading to the erroneous range of [0, ∞). The relationship between x and y is defined by the equation x² + y² = 9, indicating that y must also be limited by the maximum value of 3 when x is at its minimum or maximum. Therefore, the correct range of the function is [0, 3], as y cannot exceed this limit based on the domain of x. Understanding the connection between x and y is crucial for accurately determining the range of such functions.
arham_jain_hsr
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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...
 
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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##.
 
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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?
 
PeroK said:
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.

PeroK said:
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.
 
arham_jain_hsr said:
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.
arham_jain_hsr said:
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.
 
arham_jain_hsr said:
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.
 
PeroK said:
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?
 
Maybe if you set ##y=f(x)##, square both sides, you'll obtain a revognizable expression that will help you determine the range.
 
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arham_jain_hsr said:
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##.
 
  • #10
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)##.
 
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arham_jain_hsr said:
A circle at origin with radius 3 units?
Precisely.
 
  • #14
WWGD said:
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].
 
  • #15
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?
 
  • #16
arham_jain_hsr said:
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]##.
 
  • #18
arham_jain_hsr said:
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.
PeroK said:
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##.
 
  • #19
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 :)
 
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  • #20
arham_jain_hsr said:
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.
arham_jain_hsr 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.
 
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  • #21
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##.
 
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  • #22
Mark44 said:
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?
 
  • #23
arham_jain_hsr said:
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:
arham_jain_hsr 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##.
 
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  • #24
Mark44 said:
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.
 
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