Question

Calculus!!

Answer 1

For the function f(x) = x^2 -4x + 5, x_> 2, what is d/dx(f^-1(x))?

Answer 2

f'(1) you mean?
because f'(x)=2x-4
and all you have to do next is to plug in the number on for x

Answer 3

No. I mean f^(-1), indicating that it's an inverse function.

Answer 4

No, ^-1 means it's an exponent. The 1 doesn't isn't a value we input, it's just a simple indication that the function is meant to be an inverse function.

Answer 5

You need to find the inverse function. Say
\(y=x^2-4x+5\)
Now you just find the function as a form \(x= ___\) where ___ is a function in terms of \(y\). It is not a 1-1 function. So you'll get two solutions for \(x\). But you have a constraint \(x\ge 2\) which will likely tell you which solution to take.

Answer 6

For the inverse, I got: 2+/-sqrt(x-1).

Answer 7

actually the given function is not inyective, so its inverse does not exist.

Answer 8

The derivative I got was 1/2(sqrt(x-1) (this is when I use 2+sqrt(x-1). Among my choices in the review paper, it doesn't show that is why I am confused.

@helder_edwin, inyective? It shows that there is indeed an verse according to the choices I have to choose from.

Answer 9

oh okay im sorry
f^-1(x)=4+sqr (16−20+4x)/2

Answer 10

@helder_edwin Yes it's not injective, but the domain is restricted so that it works.

Answer 11

right! I did not see that. with the domain restricted the function is inyective. and its inverse is as @sensuelle1985 wrote
\[\large f^{-1}(x)=2+\sqrt{x-1} \]

Answer 12

Yes, that is what I got. Anyways, we are to then find the derivative of the inverse and I got 1/2(sqrt(x-1) which isn't an option. :/

Answer 13

Is here that we apply the x_>2 constraint?

Answer 14

now u can use the inverse theorem
\[\large (g^{-1})'(b)=\frac{1}{g'(g^{-1}(b))} \]

Answer 15

Oh lord, I have never even heard of the inverse theorem lol.

Answer 16

then forget it. u now have an algebraic expresion for \(f^{-1}\). apply to it the derivative rules u know.

(sorry for having brought that up.)

Answer 17

the constraint \(x\geq2\) is for the domain of \(f\), not for \(f^{-1}\).

Answer 18

That's interesting. I hadn't heard of the inverse theorem either. Thanks. So you compute the inverse function, then take the derivative of the original function, and compute f' of f^-1(x)? Why would this be useful? If it is easier to differentiate the original function rather than it's inverse?

Answer 19

u can use this theorem when there is no way to compute \(f^{-1}\) explicitly. which is not this case.

Answer 20

I don't think you take the derivative of the original function. You take the derivative of the inverse function. @ mathbrz At least that's what I thought.

Answer 21

For your problem yes, but I was asking about inverse theorem. Sorry to hijack your thread.

Answer 22

Oh, lol. All is good.

Answer 23

http://www.wolframalpha.com/input/?i=differentiate+2+%2B+sqrt%28x-1%29+and+2-sqrt%28x-1%29

Answer 24

Not that you needed help. You got this. Lol. Laters.

Answer 25

@mathbrz, that was the same answer I got except that choice wasn't given. I emailed my teacher and she told me it was 1/2y-4. 0.o I have no idea how this answer came to be but thanks for your help everybody!

Answer 26

Let g(x) be the inverse of f(x), then
\[
g(x)=\sqrt{x-1}+2
\]
All we are asked to is to compute g'(x)
\[

g'(x)=\frac{1}{2 \sqrt{x-1}}
\]
We are done.