Question

Derivatives of Inverse Functions

Answer 1

like arcisin , arccos and arctan?

Answer 2

The question is on that attachment

Answer 3

No it is not trigonometric functions

Answer 4

I'm actually doing the same thing let me see if I can get someone to help you

Answer 5

@zepdrix

Answer 6

can you find f-1(x)?

Answer 7

I could but I'm not sure if the x≥2 will affect the answer

Answer 8

the x >=2 is a restriction on the domain.

Answer 9

Yah so will that effect the outcome of the inverse function?

Answer 10

no except taht the restriction will also apply to the domain.

Answer 11

Okay then I will find the inverse function right now

Answer 12

Since this is quadriatic there will be 2 right?

Answer 13

yes

Answer 14

\[\frac{ 4+\sqrt{4x-4} }{ 2 }\] and \[\frac{ 4-\sqrt{4x-4} }{ 2 }\]

Answer 15

which simplifies to 2 +/- sqrt(x-1)

Answer 16

sorry gotta go right now.

Answer 17

Okay thanx bye

Answer 18

So you used Quadratic Function to find your inverse?
That's fine I suppose,
completing the square would have worked just as well :)

So you have determined that: \(\rm f^{-1}(x)=2\pm\sqrt{x-1}\)
Ya?

Answer 19

Should I put up my answer choices?
and yes

Answer 20

We're going to get `only plus` or `only minus`.
That's what the domain restriction was doing for us.
It was guaranteeing that our inverse would be a `function`.
So they were making a restriction to force f(x) to be one-to-one.

Answer 21

So which one, plus or minus. umm

Answer 22

Imma go with plus

Answer 23

Hmm I can't figure out how to get the correct root algebraically...
brain not working...
but if you simply look at the graph of the function,