Question

Find the range of f(x)=sqrt(9-x^2) algebraically.

Answer 1

The domain is (-infinity, 3]
But I only know how to find the range by graphing, not algebraically.

Answer 2

start with the domain and work it into the function

Answer 3

How would I do that?

Answer 4

-inf < x <= 3

since y = f(x), convert x into the f(x)

square, negate, sqrt

Answer 5

What do you mean convert x into the f(x)?

Answer 6

If I were your cat, I would lick your domain off your paper.

Answer 7

start with the domain, this < x < that

then convert x into f(x) by the appropriate transformations

Answer 8

the range is some < f(x) < thing

Answer 9

unless you told me you wanted to include some complex numbers and not others

Answer 10

is it the notation f(x) that you dont understand? it simply means a function of x, which you already have defined

Answer 11

So would it be -inf < f(x) < 3?

Answer 12

yeah, clean up your domain

9-x^2 >= 0

9 >= x^2

x <= 3 and x >= -3

Answer 13

My teacher makes us put it in interval notation, sorry.

Answer 14

im trying to rush something in my head ....

Answer 15

x=0 works, but any x^2 bigger then 9 goes negative ...

Answer 16

so -3 to 3 is the domain, agreed?

Answer 17

Wait, so would the domain be (-inf, -3] U [3, inf)?

Answer 18

Yes, what is that in interval notation so I can write that down?

Answer 19

9 - (inf)^2 = -inf

Answer 20

anything bigger than 9 goes negatice

Answer 21

I am so confused lol I have no idea what you are doing.

Answer 22

can we have a negative number under the radical?

Answer 23

My teacher never taught us this and I got it wrong on the test so I'm trying to figure out how to do everything I got wrong for future tests.
No you cannot or else it will be imaginary.

Answer 24

then we have a postive number, 9 and a number that is subtracted from it,

do you see that when we subtract something greater than 9, from 9, we go negative?

Answer 25

Yes. I see that.

Answer 26

then the only values for the domain are from -3 to 3

Answer 27

D = [-3,3]

not -inf to 3

Answer 28

I figured it out!

Answer 29

so we have the domain:

-3 <= x <= 3

we have a function defined: f(x) = sqrt(9-x^2)

the range is defined for some interval: c <= f(x) <= d

if we take the domain, and work x into f(x) then the interval itslef is worked into the range

Answer 30

Domain:
9-x^2 >= 0
-x^2 >= 9
x^2 <= 9
x <= +/- 3
-3 < x < 3

Answer 31

Interval notation would be [-3, 3]

Answer 32

yes

Answer 33

Range:
x=sqrt(9-y^2)
9-y^2 >= 0
9 >= y^2
y <= +/- 3
-3 <= x <= 3
[-3, 3]

Answer 34

now, we may need to do some mathical trick to make this work smoothly

would you agree that: -3 <= x <= 3 is that same as |x| <= 3 ??

Answer 35

that is not the range

Answer 36

Ugh. I got half of it now though lol.

Answer 37

I would agree that |x| <= 3.

Answer 38

lets transform x into f(x)

if we try this:

-3 <= x <= 3, square it all

9 <= x^2 <= 9, negate it all (and signs flip when we multiply by -1)

-9 >= -x^2 >= -9, add 9

0 >= 9-x^2 >= 0, and sqrt

0 >= sqrt(9-x^2) >= 0

which doesnt really make a lot of sense in this form, which is why writing the domain as: |x| <= 3 might work better

Answer 39

|x| <= 3

(|x|)^2 <= 9, but (|x|)^2 is equal to x^2

x^2 <= 9, negate and flip the sign

-9 <= -x^2, add 9

0 <= 9-x^2, and sqrt

0 <= sqrt(9-x^2)

i cant seem to algebra the top at the moment, its max is 3

0 <= sqrt(9-x^2) <= 3

Answer 40

y = sqrt(9-x^2)

y^2 = 9-x^2

x^2 + y^2 = 9 is a circle of radius 3 centered at the origin which gives us a max of +3

Answer 41

How did you find the max?
Is it because that is the square root of 9?

Answer 42

i found it by memory, the equation is the top half of a circle of radius 3

Answer 43

So x^2 + y^2 makes a circle?

Answer 44

yes

Answer 45

So 0 <= y <= 3
Interval notation making it [0, 3]

Answer 46

yes, but how to find the 3 eludes me algebraically

Answer 47

Well, I'm more knowledgeable about this subject than I was before, so it's a start.
Thank you anyways!

Answer 48

good luck with it :)