Question

For the function f(x)=x^2-4x+5, x is greater than or equal to 2, which is equal to d/dx (f^(-1)(x))

Answer 1

1/(2y-4) where x and y are related by the equation (satisfy the equation) x=y^2-4y+5, x is greater than or equal to 1

2y-4, where x and y are related by the equation y=x^2-4x+5, x is greater than or equal to 2

1/(2x-4), for x is greater than or equal to 1

1/(2x-4), for x is greater than or equal to 2

1/(2y-4), where x and y are related by the equation y=x^2 -4x +5, x is greater than or equal to 2

Answer 2

ok you need to know how to solve a quadratic equation
I prefer completing the square when it comes to inverse of a quad
\[y=x^2-4x+5 \\ y=x^2-4x+4+1 \\ y-1=x^2-4x+4 \]
solve for x
first complete the square

Answer 3

x = sqrt(y-1) +2

Answer 4

right
\[y-1=(x-2)^2 \\ x>2 \text{ \implies } x-2>0 \\ \text{ so } \sqrt{y-1}=x-2 \\ 2+\sqrt{y-1}=x \\ f^{-1}(x)=\sqrt{x-1}+2 \]

Answer 5

now differentiate

Answer 6

(d/dx) sqrt(x-1) +2

Answer 7

correct?

Answer 8

\[\frac{d}{dx}(f^{-1}(x))=\frac{d}{dx}(\sqrt{x-1}+2)\]

Answer 9

1/f = 1/2(sqrt(x-1))

Answer 10

you mean the derivative right?

Answer 11

I don't know where 1/f comes from

Answer 12

Yes

Answer 13

Inverse is not 1/f, a lot of people confuse that

Answer 14

\[f^{-1}(x) \neq \frac{ 1 }{ f(x) }\]

Answer 15

Ah I see.

Answer 16

\[x=2+\sqrt{y-1} \\ \text{ we are already have } x \ge 2 \text{ and we also have that we need } y \ge 1 \]

\[\frac{d}{dx} (f^{-1}(x))=\frac{1}{2 \sqrt{x-1}} \text{ where } x>1\]

Answer 17

Great explanations @freckles

Answer 18

thanks kindly

Answer 19

Is that the final step because that is not one of my possible answer choices.

Answer 20

we switched x and y right?
think we had y=x^2-4x+5

but here x is what?
x=y^2-4y+5

Answer 21

if you replace x with that what do you have

Answer 22

\[\frac{1}{2 \sqrt{x-1}}=\frac{1}{2 \sqrt{y^2-4y+5-1}}=?\]

Answer 23

remember since we switched x and y
we have x>1
and y>2 so y-2>0

Answer 24

you will need that y-2>0 to simplify

Answer 25

So if we have x =y^2-4y+5 where x is greater than or equal to 1. We get the answer:1/(2y-4) where x and y are related by the equation (satisfy the equation) x=y^2-4y+5, x is greater than or equal to 1

Answer 26

close \[\frac{1}{2 \sqrt{x-1}}=\frac{1}{2 \sqrt{y^2-4y+4+1-1}}=\frac{1}{2 \sqrt{(y-2)^2}} =\frac{1}{2 (y-2)} \text{ since } y>2 \]
and x>1 not equal to 1 since you can't plug 1 into that one expression
thought your question asked for something in terms of x not y

Answer 27

so again you have to change the variables around again :p

Answer 28

just interchange them
replace y with x

Answer 29

So which answer choice would you say is right?

Answer 30

don't forget to change the one in the inequality too

Answer 31

the 4th option is what I'm implying

Answer 32

Wow that question was confusing. I'm going to spend some time looking over this and see if I can understand it better. Thanks for all your help!

Answer 33

We did a lot of back and forth with the variables

Answer 34

there was something else we could have looked at

Answer 35

\[(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}\]
do you know this formula?

Answer 36

\[x^2-4x+5=y \\ \ (x-2)^2+1=y \\ (x-2)^2=y-1 \\ x-2=\sqrt{y-1} \text{ since } x \ge 2 \\ \text{ this also says } y \ge 1 \\ x=\sqrt{y-1}+2 \\ \text{ interchange } \\ y=\sqrt{x-1}+2 , y \ge 2 , x \ge 1 \\ f^{-1}(x)=\sqrt{x-1}+2 \\ \text{ now } f(x)=x^2-4x+5 \\ \text{ so } f'(x)=2x-4 \\ f'(f^{-1}(x))=2(\sqrt{x-1}+2)-4 =2 \sqrt{x-1} \\ \text{ so } (f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}=\frac{1}{2 \sqrt{x-1}}, x>1 , y>2 \]

you know what I'm kinda confused about your choices again

Answer 37

@rational what do you think of these choices?

Answer 38

or @Astrophysics

Answer 39

maybe you are right the first time
i thought wasn't it right because we didn't have a function of x

Answer 40

I always end up with deriving the derivative of inverse \[f^{-1}(f(x)) = x \\~\\ [f^{-1}(f(x))]' f'(x)= 1\\~\\ [f^{-1}(f(x))]' = \dfrac{1}{f'(x)} \]

Answer 41

but that works nicely for a specific value of x i think
what you did is exactly is what is asked by the question i think as they are asking for x >= 2

Answer 42

@NathanJHW the more I think of the 1st one is right
I think I'm going to say is right because it it telling you where x=y^2-4y+5
like it is telling you x-1=(y-2)^2
or x=(y-2)^2+1

Answer 43

still I think these choices are extremely odd :p

Answer 44

lol I made it harder that it should be

Answer 45

x = y^2-4y+5

dx/dy = 2y - 4

dy/dx = 1/(2y-4)

we still need to plugin the value of inverse function y here

Answer 46

like they didn't actually solve for the inverse though is the crazy thing

Answer 47

i dont think so, solving the inverse function is still required

Answer 48

I did more work than needed

Answer 49

1/(2y-4) where x and y are related by the equation (satisfy the equation) x=y^2-4y+5, x is greater than or equal to 1


is the first choice
this choice did not go as far as solving for the y
given x=y^2-4y+5

Answer 50

oh ok they gave options gotcha :)

Answer 51

thanks @rational for your workings of the problem :)