# Math Help - 3 intercepts between f(x) and inverse??

1. ## 3 intercepts between f(x) and inverse??

Something I find really odd.... A question provides us with a square root function, and asks us to sketch it and its inverse. Strangely enough, f(x) and its inverse intersect not only along y=x but at two other points.

My question is, simply, how do you figure out how many intersections there will be between a function and its inverse? Even my teacher was perplexed.

2. Originally Posted by freswood
Something I find really odd.... A question provides us with a square root function, and asks us to sketch it and its inverse. Strangely enough, f(x) and its inverse intersect not only along y=x but at two other points.

My question is, simply, how do you figure out how many intersections there will be between a function and its inverse? Even my teacher was perplexed.
You have, $f(x)=\sqrt{x}$ since this is a bijective function it have an inverse, $y=x^2 \mbox{ for } x\geq 0$. To find intersection points you need to consider,
$\sqrt{x}=x^2$ for $x\geq 0$
Square both sides,
$x=x^4$
Thus,
$x^4-x=0$
Thus,
$x(x^3-1)=0$
Thus,
$x=0$ and $x^3-1=0$
All the real solutions are,
$x=0,1$

I do not see your problem?

3. In the example they gave, it was:

f(x) = 4 - 2[root](2x+6)

We've always been taught that a function and its inverse will intersect on the line y=x, but never any mention of anything else.

4. Originally Posted by freswood
We've always been taught that a function and its inverse will intersect on the line y=x, but never any mention of anything else.
What is that supposed to mean?!?!
A function and its inverse are line relfections in line $y=x$ that is probably what you mean. Not that they intersect at all point on $y=x$

5. Originally Posted by ThePerfectHacker
What is that supposed to mean?!?!
A function and its inverse are line relfections in line $y=x$ that is probably what you mean. Not that they intersect at all point on $y=x$
In order for a graph to intersect its inverse we have to have the following situation: Let y = f(x) be the original graph. The inverse function is represented by x = f(y), or y = g(x). The only way these will intersect is if there is a point (x,y) in the original graph and a point (y,x) in the inverse graph such that (x,y)=(y,x), ie. x = y. Thus all such intersection points will be on the line y = x.

(BTW: The graph of a square root function is essentially the graph of part of a parabola either opening to the right or left. So I am going to refer to your square root function as a parabola. In the following paragraph, then, the "parabola" mentioned can either be your square root function, or its inverse.)

Now, we are looking for the set of intersection points of a parabola, its inverse, and the line y = x. A parabola can intersect with a line in AT MOST 2 points. These are the same two points that the inverse function will cross the line y = x at. So your solution set of the crossing points of your parabola and its inverse is at most 2 points, not 3.

I must therefore conclude that something was either solved or graphed incorrectly for you to have 3 points.

-Dan

6. Originally Posted by freswood
Something I find really odd.... A question provides us with a square root function, and asks us to sketch it and its inverse. Strangely enough, f(x) and its inverse intersect not only along y=x but at two other points.

My question is, simply, how do you figure out how many intersections there will be between a function and its inverse? Even my teacher was perplexed.
Hello,

in addition to all posts I only want to point out, that you have to look thoroughly at the domain and range of a function and its inverse.

$f(x)=4-\sqrt{2x+6}\ \wedge \ f(x) \leq 4$ you'll get the inverse:

$g(x)=\frac{1}{8} x^2-x-1\ \wedge \ x\leq4$

If you take g without the restriction you'll get indeed 3 intercepting points.

Greetings

EB

7. Originally Posted by freswood
Something I find really odd.... A question provides us with a square root function, and asks us to sketch it and its inverse. Strangely enough, f(x) and its inverse intersect not only along y=x but at two other points....
Hello,

it's me again.

I've attached a diagram, which shows that your teacher is right.

Greetings

EB

8. Originally Posted by earboth
Hello,

it's me again.

I've attached a diagram, which shows that your teacher is right.

Greetings

EB
Thanks for clarifying it. So how would I know (without using a calculator) whether a graph intersects its inverse more than once? Our end of year exam has a part with no calc.

9. Originally Posted by freswood
Thanks for clarifying it. So how would I know (without using a calculator) whether a graph intersects its inverse more than once? Our end of year exam has a part with no calc.
I don't know if this is of any help, but the number of points of intersection
of a function $f$ and its inverse $f^{-1}$ is equal to the number of roots of
$f(f(x))=x$ where $x$ is in the domain of both $f$ and $f^{-1}$.

This is of course equivalent to saying it is equal to the number of roots of
$f(x)=f^{-1}(x)$ in the intersection of the two domains, so its not saying
anything new really.

RonL

10. Originally Posted by earboth
Hello,

it's me again.

I've attached a diagram, which shows that your teacher is right.

Greetings

EB
My apologies. I was obviously only considering symmetric points!

-Dan