# math help1

posted by .

Okay I am a little bit confused about one math question can you please
Help me out?
There is two parts but I already finished part a just stuck on part b

Okay so part a. draw the reflection of f(x) = (square root X-2) in the
Line y=X

Okay I understand that but when it comes to part be I am totally
Clueless.

Okay here is the question!
Write the equation of the reflection in the from y=g(x)

I believe the equation is (square root x+2) +2 but x is equal too/
Greater
Than 2 and y is equal too/ greater than 0 so what do I do with this
Information?

No, the reflection of a function across the line y=x is the inverse of the function.
You have y=f(x)=sqrt(x-2). The inverse of f(x) is not (square root x+2) +2
What we want for part b is g(x)=f^-1(x), the inverse function of x.

so that would be?????

Your original f(x)=sqrt(x-2). If you let y=sqrt(x-2) we want to find the inverse of this. There are a couple way to do this. If you square both sides you get
y^2=x-2 now add two to both sides to get
y^2+2=x
now we would interchange the variables to get
y=x^2 + 2
You can verify that it is the inverse by composing the first function with the second to get
y=sqt((x^2 + 2) -2)=sqrt(x^2) = |x|
Since we're only concerned with x=>2 for f(x), we're only interested in x=>0 for g(x). Thus we can ignore the absolute value above and we have y=x. This is what we should get if the functions are inverses.

so what you are saying is that the equation is g(x)=x^2+2

Yes, I believe that's what I was saying (I didn't check) and that's what I think your original question was asking (I did check that.)
You should verify that f(g(x))=x and g(f(x))=y as they should for inverses.

so what you are saying is that the equation is g(x)=x^2+2

• math help1 -

1.1 questions no 2

## Similar Questions

2sqrt24a^3b^2*5sqrt6ab I'm guessing that the first sqrt sign is over 24a^3b^2 and the second sign is over 6ab?

(-2x^2y)^3*(5xy^3)^2=(-2x^2y)^3 = -8 x^6 y^3 *Please check my work. What did you do with the (5xy^3)^2 part. The first part you did ok. But you can't just throw away part of the equation. I don't get that part.What do I do there.?
3. ### physics

we just did an practical on newton's second law and we just have to do a report on it.... basically there are two parts to it, first part is when the overall mass of the system remains constant and the second part is when the force …
4. ### Math (Probability)

I've worked out the first part of the question, just need help with the second part which is: The spinner has three sections. While no two are the same size, one of the sections is half the size of another. • You are more likely …
5. ### Math Algebra

kassie an kesya are dividing \$39 into two parts in order that the sum of 2/3 of one part and 3/4 of the other part is \$28. What are the parts?
6. ### Reiny!!!! math help

We understand how to do the questions now and we are really glad n thankful, but the graphing portion of it is confusing. Continuation of the function. The teacher wants us to plot the points from the different parts of the questions …
7. ### Derivatives

Find f'(x): (abs(((x^2)*((3x+2) ^(1/3)))/((2x-3)^3)) So I thought I should split it into two parts: A and B. A) (x^2)/((2x-3)^2) B) ((3x+2)^(1/3))/(2x-3) My plan was to differentiate each part and then multiply them together, since …