Domain and range of multivariable functions

  • #1

Homework Statement


Specify the domain and range of f(x, y) = arccos(y − x2). Indicate whether the domain is (i)
open or closed, and (ii) bounded or unbounded. Give a clear reason in each case.


Homework Equations




The Attempt at a Solution



y-x2 >= -1
y >= x2 -1

y-x2 <= 1
y <= x2 +1

I sketched it and found the region to be in between the two parabolas
Range:
[-1;infinity)

Domain:
[1;infinity) U [-1;-infinity)

I don't know if those are correct but I got them from the sketch

I don't know what (i) and (ii) means
 

Answers and Replies

  • #2
STEMucator
Homework Helper
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Recall from real single variate calculus the domain of the ##\text{arccos}(x)## function: ##x \in [-1, 1]##. Notice this is the range of the ##\text{cos}(x)## function.

So if ##D = \{ x \in \mathbb{R} \space | \space -1 \leq x \leq 1 \}##, the range can be deduced as ##R = \{ y \in \mathbb{R} \space | \space 0 \leq y \leq \pi \}##.

Most of this translates over to real multivariate calculus. That is, you require the domain to satisfy:

$$(y - x^2) \in [-1, 1]$$

So ##D = \{ (x, y) \in \mathbb{R^2} \space | \space -1 \leq y - x^2 \leq 1 \}##.

Can you deduce the range?

For (i) and (ii), what does it mean when the domain is open/closed, bounded/unbounded?
 
  • #3
Recall from real single variate calculus the domain of the ##\text{arccos}(x)## function: ##x \in [-1, 1]##. Notice this is the range of the ##\text{cos}(x)## function.

So if ##D = \{ x \in \mathbb{R} \space | \space -1 \leq x \leq 1 \}##, the range can be deduced as ##R = \{ y \in \mathbb{R} \space | \space 0 \leq y \leq \pi \}##.

Most of this translates over to real multivariate calculus. That is, you require the domain to satisfy:

$$(y - x^2) \in [-1, 1]$$

So ##D = \{ (x, y) \in \mathbb{R^2} \space | \space -1 \leq y - x^2 \leq 1 \}##.

Can you deduce the range?

For (i) and (ii), what does it mean when the domain is open/closed, bounded/unbounded?

Is the range 0 <= z <= 180 ?, because if y-x^2 = 1, then the min value of z would be 0 and if y-x^2 had to equal -1 then the max value of z would be 180 so the range would be in between 0 and 180
 
  • #4
35,128
6,873
Is the range 0 <= z <= 180 ?
You should be thinking in terms of real numbers (i.e., radians), not degrees. The range of the arccosine function is [0, ##\pi##].
TheRedDevil18 said:
, because if y-x^2 = 1, then the min value of z would be 0 and if y-x^2 had to equal -1 then the max value of z would be 180 so the range would be in between 0 and 180.
 
  • #5
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,093
1,672
Domain:
[1;infinity) U [-1;-infinity)
I'm not sure if you were trying to specify a two-dimensional region here, but what you wrote corresponds to the union of two pieces of the number line.

The way you analyzed it graphically is fine. The notation you would use to describe that region is what Zondrina wrote.

I don't know what (i) and (ii) means
In math, it's crucial to know the precise definition of various terms as well as what they intuitively mean. If you don't know what a term means, you should look it up — that's the least you can do. If you already did this, you should have said so. If you can't make complete sense of the definition after reading about it, at least you'll have some idea of what it means. You can then ask a specific question about what's confusing you.
 

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