methods help

[hard]

find the maximal dominan. Thing is, i don't think there's a q in the book that looks like this.

The range of the function with the equation i got (2,infinte) but the answer says otherwise.

the inverse of

[TrueTears]

do you mean ?

if so just remember you can not divide by 0

so



so your domain is R\{}

if you meant

then its just a straight line, the maximal domain is R

[TrueTears]

To find the range of

Sketch the graph, it is a hyperbola with domain x>0 so range is just

[hard]

do you mean ?

if so just remember you can not divide by 0

so



so your domain is R\{}

=/= means not equal to, I don't know how to do it on latex lols

if you meant

then its just a straight line, the maximal domain is R

the former is correct but thing is, in the multiple choice there is no answer with R\{}

[EvangelionZeta]

The second one is a hyperbola, so sketch it first to see what it looks like.

Now, note that there are two halves.  There are asymptotes at x=0 and y=1, and since we're only interested in the upper half of the graph (the one that is on the right side of x=0), we only want everything that's within the curve which looks like an L.

Therefore the range is (1, infinite).



For the last one, make it into turning point form.

So

f(x)=(x-1)^2-1

Replace x with y.

x=(y-1)^2-1
x+1=(y-1)^2
(x+1)^[1/2]=y-1

Inverse f(x)=(x+1)^[1/2]+1

[EvangelionZeta]

do you mean ?

if so just remember you can not divide by 0

so



so your domain is R\{}

=/= means not equal to, I don't know how to do it on latex lols

if you meant

then its just a straight line, the maximal domain is R

the former is correct but thing is, in the multiple choice there is no answer with R\{}

Could you list the multiple choice answers that they give?

[hard]

do you mean ?

if so just remember you can not divide by 0

so



so your domain is R\{}

=/= means not equal to, I don't know how to do it on latex lols

if you meant

then its just a straight line, the maximal domain is R

the former is correct but thing is, in the multiple choice there is no answer with R\{}

Could you list the multiple choice answers that they give?

[3, infinite)
(-infinite, 3]
(0,infinite)
(3, infinite)
(-infinite, 3)

[EvangelionZeta]

The only thing which would make sense is if the (3-x) was all under the root sign (or ^[1/2]).   

In that case, 3-x must>0, since you can't divide by 0 or have a negative root.

Therefore, 3-x>0
3>x
x=(negative infinite, 3)

[TrueTears]

The only thing which would make sense is if the (3-x) was all under the root sign (or ^[1/2]).   

In that case, 3-x must>0, since you can't divide by 0 or have a negative root.

Therefore, 3-x>0
3>x
x=(negative infinite, 3)

yeah is the question meant to have square root ALL around the 3-x?

[hard]

The only thing which would make sense is if the (3-x) was all under the root sign (or ^[1/2]).   

In that case, 3-x must>0, since you can't divide by 0 or have a negative root.

Therefore, 3-x>0
3>x
x=(negative infinite, 3)

yeah is the question meant to have square root ALL around the 3-x?

yeee

[TrueTears]

anything under a square root must be larger or equal to 0

so



so

[hard]

anything under a square root must be larger or equal to 0

so



so

ahhh yee thanks for that.

[EvangelionZeta]

It's not (-infinite, 3], since the initial equation is 1/[(3-x)^(1/2)].  Root of 0 is 0, and 1/0 is undefined.

(-infinite, 3) is the answer...

[TrueTears]

It's not (-infinite, 3], since the initial equation is 1/[(3-x)^(1/2)].  Root of 0 is 0, and 1/0 is undefined.

(-infinite, 3) is the answer...

sorry, my bad.

[EvangelionZeta]

Just shows even the best of us make mistakes.  =p