Intersection of Forward and Inverse Function (1 Viewer)

laters · Apr 8, 2015

Hi,

for some exam questions I have seen they might give an equation (say f(x)=e^x - 4) and ask why the x coordinate of any intersection points of f(x) and its inverse satisfy an equation (e^x - x - 4 = 0) which obviously requires you to equate f(x)=x.

But in the textbook I have seen a question y=-x^3 which intersects with its inverse on the line y=-x.

So is there a general rule or something I need to know? Or should I be actually equating the forward with the inverse all the time, or draw a graph to make sure?

turntaker · Apr 8, 2015

I'd equate them and/or graph to make sure.

braintic · Apr 8, 2015

A function and its inverse don't have to intersect on y=x or on y=-x.

confusedcourse · Aug 16, 2015

braintic said:
A function and its inverse don't have to intersect on y=x or on y=-x.

Why not? (unless their asymptotes are y = x, -x)

InteGrand · Aug 16, 2015

confusedcourse said:
Why not? (unless their asymptotes are y = x, -x)

$If the graph of $y=f(x)$ happens to pass through the points $A(a,b)$ and $B(b,a)$ with $|a|\neq |b|$, then the graph of $y=f^{-1}(x)$ will also pass through the points $A$ and $B$, and these aren't on the lines $y=\pm x$.$

braintic · Aug 16, 2015

confusedcourse said:
Why not? (unless their asymptotes are y = x, -x)

My wording does not precisely reflect what I was trying to say.

It should be:
The points of intersection of a function and its inverse don't have to lie on y=x or y=-x.

leehuan · Aug 17, 2015

InteGrand said:
$If the graph of $y=f(x)$ happens to pass through the points $A(a,b)$ and $B(b,a)$ with $|a|\neq |b|$, then the graph of $y=f^{-1}(x)$ will also pass through the points $A$ and $B$, and these aren't on the lines $y=\pm x$.$

braintic said:
My wording does not precisely reflect what I was trying to say.

It should be:
The points of intersection of a function and its inverse don't have to lie on y=x or y=-x.

This only makes sense to me logically. Can someone please care to provide an example?

InteGrand · Aug 17, 2015

leehuan said:
This only makes sense to me logically. Can someone please care to provide an example?

$The points $A(a,b)$ and $B(b,a)$ are points that are reflections of each other about the line $y=x$. A simple example would be the straight line passing through these points; the inverse would be the same line, so the function and the inverse intersect everywhere on their graphs (as they are the same), hence including on points not on $y =\pm x$.$

leehuan · Aug 17, 2015

Fair enough, I forgot about the family of curves that have inverse functions as themselves.

y=-x+b; y=c/x; y=-c/x. y=x etc.

InteGrand · Aug 17, 2015

leehuan said:
Fair enough, I forgot about the family of curves that have inverse functions as themselves.

y=-x+b; y=c/x; y=-c/x. y=x etc.

$Other self-inverse functions include:$

$\bullet$ $f:\mathbb{R}\backslash \{-1\}\to \mathbb{R}\backslash \{-1\}, f(x) = \frac{1-x}{1+x}$

$\bullet$ $f:\mathbb{R}\backslash \{a\}\to \mathbb{R}\backslash \{a\}, f(x) = a+\frac{1}{x-a}$, where $a\in \mathbb{R}$

$$\bullet$ $f:\mathbb{R}^+\to \mathbb{R}^+, f(x) = \ln \left(\frac{\mathrm{e}^x +1}{\mathrm{e}^x -1} \right)$.$

$Such functions are called \textsl{involutions} and satisfy $f(f(x))=x$ (this equation is referred to as \textsl{Babbage's functional equation}).$

Intersection of Forward and Inverse Function (1 Viewer)

laters

Member

turntaker

Well-Known Member

braintic

Well-Known Member

confusedcourse

New Member

InteGrand

Well-Known Member

braintic

Well-Known Member

leehuan

Well-Known Member

InteGrand

Well-Known Member

leehuan

Well-Known Member

InteGrand

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)