# Domain and range of multivariable functions

## 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.

## 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

STEMucator
Homework Helper
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?

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

Mark44
Mentor
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.

vela
Staff Emeritus