Question

Find the value(s) of x for which f(x)=f^-1(x) where
f(x)=sqrt(x+2)-2/3
f^-1(x)=(x+2/3)^2-2

Answer 1

er I know f-1f(x)=x

Answer 2

as in I know how to get f^-1 but I don't know how to go on from there

Answer 3

oops sorry i made an error..! forget the abv statement..!

Answer 4

oh they have given the value of both... so just equate them!

Answer 5

\[\sqrt{x+2}-2/3 = (x+2/3) ^2 - 2\] ok?

Answer 6

yup but I'm not sure how to equate. That's the problem

Answer 7

As in I end up with x^4 or something like that and I'm not sure how to solve from there

Answer 8

yeah i see your problem..! then that is not the way to do it.. a easier way is to use the abv property that you wrote..

Answer 9

\[f(f^{-1}(x)) = x\] right? how can you rewrite this here..?

Answer 10

@wcrmelissa2001 ?

Answer 11

sorry im here

Answer 12

wait but if you were to do that then how would you explain that they are both equal?

Answer 13

yes use this info in the above eq..

Answer 14

no, as in I don't understand how the f-1f(x)=x will work in this case since it doesn't state that f-1(x)=f(x)???

Answer 15

no i meant more like:\[f(f(x)) =x\]..but even here there is a term with x^4..hmm.. let me see if i can think of a better method..otherwise we must do it the long way..

Answer 16

I tried drawing two graphs then finding their intersection too but nope

Answer 17

inverse functions are symmetric about the line y=x

if they are ever equal, it is when they meet on that line.
so for f(x) = sqrt(x+2)-⅔
i.e. y = sqrt(x+2) - ⅔
we can replace the y with x to find the spot on the y=x line where the function crosses

\[ x = \sqrt{x+2} - \frac{2}{3} \]

Answer 18

but that is the original function?

Answer 19

maybe I have to rethink this. But here is a graph from wolfram
http://www.wolframalpha.com/input/?i=plot+y%3D(x%2B%E2%85%94)%5E2+-2;+y%3D+sqrt(x%2B2)+-+%E2%85%94;+y%3D+x+for+x%3D-2+to+%2B2

Answer 20

this plot shows the symmetry better, but my idea only gets one of the solutions

Answer 21

i plotted the graph too :(

Answer 22

what course is this ? what grade ?

Answer 23

not sure... I don't take grades. Different system :D

Answer 24

how much math are you supposed to know ?

Answer 25

The brute force way is solve the quartic , which is possible , though painful by hand
but is there an elegant way ? Off hand I don't know. (though we can find the positive solution easily enough) (-1+ sqrt(57))/6

Answer 26

\[f(x)=\sqrt{x+2}-\frac{2}{3}\]

therefore \(f:[-2,\infty)\to[-2/3,\infty)\)

hence
\[f^{-1}(x)=\left(x+\frac{2}{3}\right)^2-2\]

and \(f^{-1}:[-2/3,\infty)\to[-2,\infty)\)

Answer 27

sorry i'm not sure i understand your working?

Answer 28

To me, just let them equal and solve for x. It is a mess but doable still. At the end up, you can have at most 4 values of x.

Answer 29

Let g(x) be the inverse of f(x). Then we want to solve
f(x) = g(x)
This gives a 4th order equation. Messy to solve unless you use wolfram or mathematica.

Here is one way to solve the quartic equation, in "chunks"
the idea is the 4 roots of the quartic means you can factor it into
(x-a)(x-b)(x-c)(x-d) = 0

using the idea that two of the roots lie on the line y=x
we can solve (either) equation: f(x) = x
this is a quadratic and gives two of the roots. One of them is what we want.

to find the other two roots, we divide the quartic that arises from f(x)=g(x)
by the quadratic that arises from f(x)=x

this gives us a new quadratic. in other words, if we found two roots a, and b
we are dividing the quartic by (x-a)(x-b) to get (x-c)(x-d)
which we can now solve using the quadratic formula.
Still a bit of a mess, but doable.

Answer 30

I think everyone here needs to read the definition of an inverse function again

https://en.wikipedia.org/wiki/Inverse_function#Definitions

Answer 31

Thanks everyone! :D Will go and try it out and keep trying and trying :( HAHA