Simple Algebra Problem:

[dleccord 5,000+ posts]

Find the domain and range of f(x) = sqrt{1-sqrt[1-sqrt(x)]}.

Click to read more...

 

[matrxx dude 5,000+ posts]

right

 

[dragnix 5,000+ posts]

domain=all non-negative numbers?

 

[dleccord 5,000+ posts]

domain=all non-negative numbers?

True. Up to 1. It's 0>=x>=1. But how?

Need help with the range as well. Anyone?

 

[dleccord 5,000+ posts]

right

How bout the chick in my avatar, far left?

 

[dragnix 5,000+ posts]

How bout the chick in my avatar, far left?

far right, i need her name //content.invisioncic.com/y282845/emoticons/fyi.gif.9f1f679348da7204ce960cfc74bca8e0.gif

if only i could see how the problem is actually written, the way you have it is a bit confusing.

 

[kballa422 10+ year member]

How bout the chick in my avatar, far left?

I need the name of every chick in your sig.

Also, the domain is 0
Oh, and to find the range just plug in both extremes of the domain into the problem to get your range of y values.

Someone correct me if I'm wrong.

 

[dleccord 5,000+ posts]

far right, i need her name //content.invisioncic.com/y282845/emoticons/fyi.gif.9f1f679348da7204ce960cfc74bca8e0.gif
if only i could see how the problem is actually written, the way you have it is a bit confusing.

help me find the answer and ill give you her name.

 

[bk12321 10+ year member]

pretty sure its 0

x = domain

y = range

the reason for that is the only numbers you can plug into the equation for it to work are between 0 and 1, say you want to plug in a 2...

f(2)= sqrt(1 - sqrt(1 - sqrt(2)))

you would get a negative sqrt because sqrt of 2 is 1.something and sqrt of (1 - 1.something) is negative, so x HAS to be between 1 and 0. you cant take a negative sqrt (unless you bring in imaginary numbers, and well im not getting into that).

now you plug in 0 for x and plug in 1 for x....

which gives you: f(1) = 1,

f(0) = 0

so y has to be between 0 and 1 because we plugged in both ends of the domain (x = 0, x = 1) and got those 2 answers.

basically you just want to find ALL of the possible answers that can make that equation work.

if you need any further clarification my aim is bk12321, wayyyy easier than posting on here

 

[dragnix 5,000+ posts]

range is (0,1] i believe

 

[kballa422 10+ year member]

pretty sure its 0
x = domain

y = range

I second this answer. And I tried to explain why the domain is such in above post.

Now for those names please //content.invisioncic.com/y282845/emoticons/biggrin.gif.d71a5d36fcbab170f2364c9f2e3946cb.gif

 

[dragnix 5,000+ posts]

I second this answer. And I tried to explain why the domain is such in above post.
Now for those names please //content.invisioncic.com/y282845/emoticons/biggrin.gif.d71a5d36fcbab170f2364c9f2e3946cb.gif

//content.invisioncic.com/y282845/emoticons/word.gif.64b12e39f936af3b4fff38a1c0bd0244.gif

 

[bk12321 10+ year member]

range is (0,1] i believe

pretty sure the range is [0,1] but ill double check my work:

edit:

f(0) = sqrt(1- sqrt(1- sqrt(0)))

f(0) = sqrt(1- sqrt(1- 0))

f(0) = sqrt(1- 1)

f(0) = sqrt(0) = 0

f(x) = [0,1]

0 is included because you CAN take the square root of 0, its just 0.

 

[_JaY_ 10+ year member]

um...a range is what you cook on and a domain is a place you live, idk what you guys are talkin about HAHAHAH.

 

[dragnix 5,000+ posts]

pretty sure the range is [0,1] but ill double check my work:
edit:

f(0) = sqrt(1- sqrt(1- sqrt(0)))

f(0) = sqrt(1- sqrt(1- 0))

f(0) = sqrt(1- 1)

f(0) = sqrt(0) = 0

f(x) = [0,1]

0 is included because you CAN take the square root of 0, its just 0.

yea you're right, brain fart. And where's the name //content.invisioncic.com/y282845/emoticons/mad.gif.c18f003ab0ef8a0d9c27ca78d77a6392.gif