Inverse (1 Viewer)

[lyounamu]

When you find the intersection between inverse functions, how do you come about approaching this?

For example, if I have f(x) = square root of (x+6) and f-1(x) = x^2 - 6,

I just substitutde f(x) value in for the x value in f-1(x). Would it be correct to do so?

 

[Dota55]

get one of the equations and equate it with y=x.

 

[Aerath]

Wouldn't it be easier to find where either the forward function or inverse function intersects with y = x?

Edit: Or do what Dota55 said.

 

[lyounamu]

No, that's not what I meant. I already found both inverse and the normal one.

Now, I want to find the intersection between these two graphs.

 

[conics2008]

key things to remmeber are.

f(x) domain >> f-1(x)= range
f(x) range >> F-1(x) domain

thats how you find intercepts.

 

[lyounamu]

...

What I am trying to find is the interesection between the two graphs when they have an intersection. I know how to find it but I am just not sure if I can just say

f(x) = square root of (x+6) and the inverse f-1(x) = x^2 - 6.

Therefore, substitute f(x) into the value of x in f-1(x).

Therefore, the intersection is (square root of (x-6))^2 - 6 = x.

 

[lyounamu]

Anyone care to answer my opening question?

 

[Aerath]

If I'm understanding your question correctly - I think it would be right to equate f(x) = f^-1(x), however, it's just (relatively) long and complicated. You might as well just equate f(x) = x

 

[lyounamu]

Ok, this is what i think you've got:

Two eqns, both independent of each other, but which share a special relationship (being the inverse of each other). So essentially:

y = (x + 6)^1/2 ...... (1)
y = x^2 -6 .............. (2)

To find the pt of intersection, you'd equate the two:

(x + 6)^1/2 = x^2 -6

and you'd get:

x^4 - 12x^2 - x + 30 = 0.

But what you're asking is whether you can sub y(1) into eqn(2). (refering to post #6)
Which means you'd be subbing y(1) into x(2), do you see?

Which is incorrect.

Am i making sense?

I should have said the equation of the intersection. I seriously get the terms wrong all the time. What I wanted to find was the equation of the intersection of the two graphs.

 

[conics2008]

Here use this roots

x=-2 and x=3 ( you get them by looking at 30 )

(x+2)(X-3) = X^2-X-6

DIVIDE BY x^4 - 12x^2 - x + 30 =0

which gives u (x^2-x-6)(x^2+x-5)

there fore you can get x=-2 x=3

now work on x^2+x-5

-1+root of 21 /2 or -1-root of 21/2

basically if you want the equation. find the gradient of the curve by dy/dx and then get graident .. you use these points. make sure you get y value aswell..

remember with your graph there are only 2 which are 3 and find the one which is one of the root thingy.

 

[shadyhaze]

huh? intersection is when f(x) or f-1(x) = x
so x² - 6 = x
x² - x - 6 = 0
(x-3)(x+2)= 0
x = 3, -2
however f(-2) = root -4
therefore f(-2) is not a solution
at f(3) = root 3+6 = 3
point of intersection = (3,3)
i dont understand what other ppl are doing :S

 

[Aerath]

huh? intersection is when f(x) or f-1(x) = x

Well, that's what most of us thought, but then Lyounamu said that that wasn't what he was looking for. =\

 

[lyounamu]

Well, that's what most of us thought, but then Lyounamu said that that wasn't what he was looking for. =\

It really was a random question from some random paper. I really didn't get what it was asking but I figured that out after long consideration. Thanks. No worries.