# inverse function intersections

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• Apr 25th 2009, 09:05 PM
ella85
inverse function intersections
So theres this question in my textbook:..
Find the point of intersection between the function y= (3/((3x-4)^2)) + 6

I know you begin with

x = (3/((3x-4)^2)) + 6

x-6 = (3/((3x-4)^2))

√ (x-6)(3x-4)= √ 3

but then i just keep going round in circles and can never get the x on its own.
any help would be greatly appreciated=]
• Apr 25th 2009, 10:45 PM
pickslides
Quote:

Originally Posted by ella85
So theres this question in my textbook:..
Find the point of intersection between the function y= (3/((3x-4)^2)) + 6

I know you begin with

x = (3/((3x-4)^2)) + 6

x-6 = (3/((3x-4)^2))

√ (x-6)(3x-4)= √ 3

but then i just keep going round in circles and can never get the x on its own.
any help would be greatly appreciated=]

Hi ella85

I'm assuming you want to find the interesection of where $f(x) = f(x)^{-1}$

If so you don't have to actually find the inverse of $f(x)$, you just have to make your function $f(x) = x$ as to points of intersection of a function an its inverse are always found on the line $y = x$

Therefore simply solve

$\frac{3}{(3x-4)^2} + 6 = x$

$\frac{3}{(3x-4)^2} = x - 6$

$3 = ( x- 6)(3x-4)^2$

etc...

I think the solutions are $x = 9, \frac{\sqrt{3}-4}{3}$
• Apr 25th 2009, 11:05 PM
mr fantastic
Quote:

Originally Posted by pickslides
Hi ella85

I'm assuming you want to find the interesection of where $f(x) = f(x)^{-1}$

If so you don't have to actually find the inverse of $f(x)$, you just have to make your function $f(x) = x$ as to points of intersection of a function an its inverse are always found on the line $y = x$

In this case that's true. But in general the points of intersection of a function and its inverse can also lie on lines of the form $y = - x + c$ where $c$ is a constant.

eg. Two of the three intersection points of the function

$f: [0, \, + \infty) \longrightarrow R, ~ f(x) = -x^2 + 1$

and its inverse function lie on the line $y = -x + 1$.
• Apr 26th 2009, 12:27 AM
ella85
thanks for replying=]
sorry, i dont think i explained what im having trouble with properly.
I understand how to find the intersection of a function and its inverse, but I cant solve this particular equation

3x-4 = √(3/(x=6))

i need to make x the subject, but when i square it to get rid of the square root, it still doesnt work..
• Apr 26th 2009, 04:29 AM
mr fantastic
Quote:

Originally Posted by ella85
So theres this question in my textbook:..
Find the point of intersection between the function y= (3/((3x-4)^2)) + 6

I know you begin with

x = (3/((3x-4)^2)) + 6

x-6 = (3/((3x-4)^2))

√ (x-6)(3x-4)= √ 3

but then i just keep going round in circles and can never get the x on its own.
any help would be greatly appreciated=]

If $f(x) = \frac{3}{(3x - 4)^2} + 6$ then an inverse function does not exist. This is because $y = f(x)$ is not one-to-one. So the first thing you have to do is restrict the domain of $y = f(x)$ so that is is one-to-one.

One such restriction might be $\left( \frac{3}{4}, ~ + \infty\right]$. Then $f^{-1}(x) = -\frac{1}{3} \sqrt{\frac{3}{y - 6}} + \frac{4}{3}$.

How I got this:

Solve $x = \frac{3}{(3y - 4)^2} + 6$ for $y$:

$x = \frac{3}{(3y - 4)^2} + 6$

$\Rightarrow x - 6 = \frac{3}{(3y - 4)^2}$

$\Rightarrow \frac{3}{x - 6} = (3y - 4)^2$

$\Rightarrow \pm \sqrt{\frac{3}{x - 6}} = 3y - 4$

$\Rightarrow \pm \sqrt{\frac{3}{x - 6}} + 4 = 3y$.

A simple known point on $y = f(x)$ is (1, 9). Therefore a point on $y = f^{-1}(x)$ is (9, 1). Substitute (9, 1) into $y = \pm \frac{1}{3} \sqrt{\frac{3}{y - 6}} + \frac{4}{3}$:

$1 = \pm \frac{1}{3} \sqrt{\frac{3}{9 - 6}} + \frac{4}{3}$

and so the negative root solution for the inverse function must be used to get equality.

--------------------------------------------------------------------------------------------------------------------------

If the restriction $\left(- \infty, ~ \frac{3}{4}\right]$ on the domain of $y = f(x)$ is used, then $f^{-1}(x) = \frac{1}{3} \sqrt{\frac{3}{y - 6}} + \frac{4}{3}$.
• Apr 26th 2009, 04:49 AM
ella85
I'm really sorry to keep bothering you...you're explaining it all really well
but what im trying to find is the point of intersection between
http://www.mathhelpforum.com/math-he...8a4e9ac9-1.gif and the line y=x

so i know that you use simultaneous equations and ive simplified it to the point where
x-6 = 3/((3x-4)^2)
and it says in my book that the answer is 6.015, but i can't work out how to get this answer.
thanks again.
• Apr 26th 2009, 05:22 AM
mr fantastic
Quote:

Originally Posted by ella85
I'm really sorry to keep bothering you...you're explaining it all really well
but what im trying to find is the point of intersection between
http://www.mathhelpforum.com/math-he...8a4e9ac9-1.gif and the line y=x

so i know that you use simultaneous equations and ive simplified it to the point where
x-6 = 3/((3x-4)^2)
and it says in my book that the answer is 6.015, but i can't work out how to get this answer.
thanks again.

Are you expected to get an exact answer or is an approximate answer sufficient? From the answer given, it looks like the latter. In which case you should use technology (eg. a graphics or CAS calculator) to solve $(x - 6)(3x - 4)^2 = 3$.

To get an exact answer will involve solving a cubic equation that has no simple exact real solution (which means you would have to use the Cardano formula).