Question

pls help me to find out function's domain and range. Next, describe level curves of the functions
i) f(x,y)= x-y
ii) f(x,y) = x^2+y^2
iii) f(x,y) = x^2-y^2 in square root
iv) f(x,y,z) = x+y+z

Answer 1

So you need to state the domain and range, as well as a description of the level curves of each function?

Answer 2

yes

Answer 3

Have you gone through any of these yet? If not, we can start with the first one.

Answer 4

ok

Answer 5

Ok so let's do the first one then?

Answer 6

So when determining the domain of any given function, we need to describe the set of all values of the independent variables. For i ), would you say that there are any restrictions on the domain?

Answer 7

i think no

Answer 8

Mmmmmk that's correct. And what's your reasoning behind that?

Answer 9

zero denominators and negatives under square roots

Answer 10

Let's consider the range now.

Answer 11

Wait. . .

Answer 12

When a real-valued function is defined along all values of the x-axis, we say that its domain is R, or, the set of all real numbers.

Answer 13

Likewise, we can ask ourselves if there are any restrictions upon the range.

Answer 14

In other words, what is the set of all values of f(x,y)

Answer 15

no

Answer 16

Then, like the domain, the range is all real numbers.

Answer 17

0 to infinity

Answer 18

I would just write something like dom f(x,y) = R and ran f(x,y) = R

Answer 19

Or, in interval notation, from (-inf, inf)

Answer 20

Do you know what notation I am referring to when I write R?

Answer 21

real number

Answer 22

Yes, all real numbers. OK. What do you think the dom and ran of ii ) is?

Answer 23

same

Answer 24

Mmmmmm, not quite.

Answer 25

The domain is the same. Both x and y can be any real number. However, what's the lower bound of the range? Is the range ever negative?

Answer 26

no..from 0 to infinity

Answer 27

Yes.

Answer 28

Oh, we forgot the level curves. Ok, real quick, then I have to go. For i ), if we let x - y = c , where c is an arbitrary constant, then we derive a level curve of the surface in three-space. Start by letting c = 0.

Answer 29

What kind of equation do you derive from doing this and along what value of f(x,y) does the image lie on?

Answer 30

functions of several variables and the graph?

Answer 31

Here, if we let x- y = 0, then it follows that y = x and so we have a line that lies along the plane f(x, y) = 0.

Answer 32

I think technically the domain is \[R\times R\] and the range is \(R\)

Answer 33

for the first one anyway...

Answer 34

I think you are right. Since there are two independent variables.

Answer 35

well functions take an input, and two different elements would not be 1 input. So we are are looking at f((x,y)) where the input is an element of some set of ordered pairs.

Answer 36

Anyway sorry to interrupt its a small distinction, and one that maybe the teacher would not care about:)

Answer 37

Yes, thank you for clarifying. I'm like thinking half between R and R X R. . .

Answer 38

I must go. Good night and good luck. I think you have the general idea as far as the dom and ran go. You can go to Pauls Online Notes or the Khan Academy for more info regarding this topic.

Answer 39

ok..tq for your kindness reply

Answer 40

pls help me to solve this Use the method of Lagrange multipliers to find the local extreme values of the function f
subject to the constraint:

We introduce the multiplier λ and solve the system ∇=∇ f g λ , g = 0.

i. f x y xy (, ) = on the ellipse 2 2 x y + = 4 1
ii. f xy x y (, ) 2 6 = −+ on the circle 2 2 x y + = 4

Answer 41

Use the method of Lagrange multipliers to find the local extreme values of the function f
subject to the constraint:

We introduce the multiplier λ and solve the system ∇f=λdelta g , g = 0.

i. f( x, y)= xy on the ellipse x^2+4y^2= 1
ii. f (x,y)=2x- y+ 6 on the circle x^2 + y^2= 4

Answer 42

okay, for (i) we know \(\nabla f=(y,x)\) whereas \(\nabla g=(2x,8y)\) so:$$y=2\lambda x,x=8\lambda y$$and thus \(y=2\lambda(8\lambda y)=16\lambda ^2y\) meaning either \((16\lambda^2-1)y=0\) so \(x=y=0\) or \(\lambda=\pm1/4\). since \(x=y=0\) does not satisfy the constraint we look substitute in for \(\lambda\) to get \(x=\pm2y\):$$x^2+4y^2=1\\(\pm2y)^2+4y^2=1\\4y^2+4y^2=1\\8y^2=1\\y=\pm\frac1{2\sqrt2}$$thus \(x=\pm1/\sqrt2\) and we get four extrema