Functions and Inverses

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January 14th 2007, 06:10 PM
Freaky-Person

Functions and Inverses

just 2. I got the rest down but these ones are driving me insane.

Find inverse of functions

1. h(x) = 2x/(x+10)

2. f(x) = x^2 + 6x + 11

Thanks.
January 14th 2007, 07:09 PM
ThePerfectHacker

Quote:

Originally Posted by Freaky-Person

1. h(x) = 2x/(x+10)

$y=\frac{2x}{x+10}$
Substitute x for y and solve,
$x=\frac{2y}{y+10}$
$x(y+10)=2y$
$xy+10x=2y$
$2y-xy=10x$
$y(2-x)=10x$
$y=\frac{10x}{2-x}$

Quote:

2. f(x) = x^2 + 6x + 11

It has no inverse.
Horizontal line test, a parabola can be passes twice or more with a horizontal line.
January 14th 2007, 07:49 PM
Freaky-Person

Quote:

It has no inverse.
Horizontal line test, a parabola can be passes twice or more with a horizontal line.

I know it stops being a function, but there should still be an inverse right? Cause you can still draw y=x and flip the original perabola along that line to get the shape, even if it's not a function.

Like the inverse of y = X^2 is y = +-sqrtX, it's still an inverse even if it's not a function....
January 14th 2007, 07:52 PM
AfterShock

Quote:

Originally Posted by ThePerfectHacker

$y=\frac{2x}{x+10}$
Substitute x for y and solve,
$x=\frac{2y}{y+10}$
$x(y+10)=2y$
$xy+10x=2y$
$2y-xy=10x$
$y(2-x)=10x$
$y=\frac{10}{2-x}$

It has no inverse.
Horizontal line test, a parabola can be passes twice or more with a horizontal line.

Minor error on #1; did not carry over the x.

Should be (10x)/(x-2)
January 14th 2007, 08:04 PM
AfterShock

Quote:

Originally Posted by Freaky-Person

I know it stops being a function, but there should still be an inverse right? Cause you can still draw y=x and flip the original perabola along that line to get the shape, even if it's not a function.

Like the inverse of y = X^2 is y = +-sqrtX, it's still an inverse even if it's not a function....

f^(-1)(x) = {(y,x) | such that y = f(x)}

f will not have an inverse if its not a one-to-one function.
January 14th 2007, 08:07 PM
Freaky-Person

Quote:

Originally Posted by AfterShock

f^(-1)(x) = {(y,x) | such that y = f(x)}

f will not have an inverse if its not a one-to-one function.

whaaaa!! Then what am I supposed to do!?! It says
For the function f(x) = x^2 + 6x + 11

a)sketch the function and its inverse on the same grid

b) Find the equation of the inverse

...:(
January 14th 2007, 08:56 PM
CaptainBlack

Quote:

Originally Posted by Freaky-Person

Find inverse of functions

2. f(x) = x^2 + 6x + 11

Put:

$y=x^2+6x+11$

then:

$x^2+6x+11-y=0$ ,

Now use the quadratic formula to get:

$x=\frac{-6 \pm \sqrt{36-4(11-y)}}{2}$

or:

$g(y)=\frac{-6 \pm \sqrt{36-4(11-y)}}{2}$ .

Which is not a function when you are looking for functions from
R to R as it fails to be single valued, but it is what you are expected to
produce, and there are interpretations under which it is a function
(such as the extension of f and g to mappings from P(R) the set of
subsets of R, to itself) .

RonL
January 15th 2007, 06:21 AM
ThePerfectHacker

Maybe he is trying to find the inverse,
$f^{-1}: f[\mathbb{R}]\to \mathbb{R}$ .
Meaning from the image of the function to the set of real numbers. In that case an inverse exists.
January 15th 2007, 02:55 PM
Freaky-Person

Kay I got it. I'll post it up for future reference or something like that.

f(x) = x^2 + 6x + 11

y= x^2 + 6x + 11

for inverse

x = y^2 + 6y + 11

x = (y^2 + 6y + 9) + 2

x - 2 = (y + 3)^2

+-sqrt(x - 2) = y + 3

+-sqrt(x - 2) - 3 = y

BTW, anyone know any sort of tutorial for that big letter CODE writing?
January 15th 2007, 03:03 PM
Freaky-Person

Quote:

Originally Posted by ThePerfectHacker

Maybe he is trying to find the inverse,
$f^{-1}: f[\mathbb{R}]\to \mathbb{R}$ .
Meaning from the image of the function to the set of real numbers. In that case an inverse exists.

I have no idea what you just said there, but it's probably it.
January 15th 2007, 03:50 PM
ThePerfectHacker

Quote:

Originally Posted by Freaky-Person

BTW, anyone know any sort of tutorial for that big letter CODE writing?

I am impressed, a teenager who actually wants to learn something! I wish more of your types existed, otherwise their philosphy is simple, "If it not on the exam we do not care, we do not need it".
http://www.mathhelpforum.com/math-he...-tutorial.html