1. ## Inverse Function help

f(x)= -√(x^2-16) ; x > or equal to 4 .......

I have to show that this function is one to one, and then i have to find the inverse function
i cant use line tests to show that this is one to one, i have to show it algebraiclly or whatever, and finding the inverse by switching variables is throwing me off i donno, please help

2. yo im stupid iono wat i was doing wrong... but just to be sure the inverse would be g(x)=sqrt(x^2+16) .... correct? but how would i show that the original is one to one on that interval? ... i mean can i just use f(a)=f(b) and solve, would that show that a=b?

argh but wait when i try to compose the inverse and the original fn both ways...f of g of x and g of f of x, i get -|x|, and then one way it doesnt just give me x, this stuff is confusing the hell out of me, please someone just post up the right answer cuz im goin nuts

3. Not sure if this is an entirely correct way of approaching the problem, but like what you said, show that if f(x) = f(y) --> x = y

Let f(x) = -√(x^2-16) for x >= 4 and f(y) = -√(y^2-16) for y >=4

If x = y, then f(x)=f(y)... so we suppose that x not = y to begin with, and hope that id f(x) = f(y), then x = y

If f(x)=f(y),

-√(x^2-16) = -√(y^2-16)
√(x^2-16) = √(y^2-16)
x^2-16 = y^2-16
x^2 = y^2

Therefore y = x or y = -x

y cannot = -x as we said that x and y are both >= 4

so y = x. Proving that the function is infact one-to-one (injective)

The second part...

Let f(x) = y = -√(x^2-16) for x >= 4

To find its inverse, we swap the x and y's, so

x = -√(y^2-16) for y >= 4
-x = √(y^2-16)
x^2 = y^2 -16
y^2 = x^2 + 16
y = √(x^2 + 16)

looking at the restriction,

y >= 4
y^2 >= 16
y^2 - 16 >= 0
√(y^2 - 16) >= 0
-√(y^2 - 16) <= 0
x <= 0

So the inverse functino g(x) = √(x^2 + 16) for x <= 0

Hope that helps.

4. ..