# Thread:

1. ## 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.

2. Originally Posted by Freaky-Person

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

3. 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....

4. 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)

5. 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.

6. 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

...

7. 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

8. 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.

9. 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?

10. 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.

11. 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
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