# Invertible functions

• Jan 24th 2010, 08:42 AM
liptonpc
Invertible functions
i try do it but only first one #_# 2 and 3 i dont know how can do #_#(Headbang)
http://www4.picturepush.com/photo/a/...40/2822122.jpg

Thanks.
• Jan 24th 2010, 08:54 AM
Jhevon
Quote:

Originally Posted by liptonpc
i try do it but only first one #_# 2 and 3 i dont know how can do #_#(Headbang)
http://www4.picturepush.com/photo/a/...40/2822122.jpg

Thanks.

remember, $f^{-1}$ undoes $f$. This means, the output becomes the input. Let $y = f^{-1}(x)$, then

$x = 3y^2 - 1$

solve for y

(Now you should notice a problem here...this function does not have an inverse. not unless the domain was restricted some way)

for (iii) $\frac fg (x) = \frac {3x^2 - 1}{x + 5}$

now just plug in $x = 1$ and evaluate
• Jan 24th 2010, 12:22 PM
liptonpc
$f^{-1}(x)=\sqrt\frac {x + 1}{3}
$

is it correct ?
• Jan 24th 2010, 12:31 PM
Jhevon
Quote:

Originally Posted by liptonpc
$f^{-1}(x)=\sqrt\frac {x + 1}{3}
$

is it correct ?

technically you need a +/- in front of that. which would make $f^{-1}$ not a function.

only bijective functions are invetible, and f isn't bijective
• Jan 24th 2010, 01:22 PM
liptonpc
sorry i dont understand #_#
so the result is ?
• Jan 24th 2010, 02:16 PM
Jhevon
Quote:

Originally Posted by liptonpc
sorry i dont understand #_#
so the result is ?

f does not pass the horizontal line test and is hence not invertible. it has no inverse. not unless the domain was restricted somehow.
• Jan 25th 2010, 12:26 PM
liptonpc
$f^{-1}(x)=\sqrt\frac {x + 1}{3}$
when i check
$f(f^{-1}(x))=x$

so why it not inverse?
• Jan 25th 2010, 12:31 PM
Jhevon
Quote:

Originally Posted by liptonpc
$f^{-1}(x)=\sqrt\frac {x + 1}{3}$
when i check
$f(f^{-1}(x))=x$

so why it not inverse?

because $f^{-1}(x) \ne \sqrt{\frac {x + 1}3}$, rather, $f^{-1} (x) = ~{\color{red} \pm}~ \sqrt {\frac {x + 1}3}$

which means $f^{-1}$ is not a function at all, since that does not pass the vertical line test.
• Jan 25th 2010, 12:41 PM
liptonpc
if $f(x)=x^2-9$
$f^{-1}(x)=\pm(x+9)^{\frac{1}2}$
and same thing it not inverse?
• Jan 25th 2010, 12:45 PM
Jhevon
Quote:

Originally Posted by liptonpc
if $f(x)=x^2-9$
$f^{-1}(x)=\pm(x+9)^{\frac{1}2}$
and same thing it not inverse?

correct.

now on the other hand, if they had restricted the domain and said, $f(x) = x^2 - 9$ for $x \ge 0$, then you could say $f^{-1}(x) = \sqrt {x + 9}$ and that's the inverse function.

we don't need the negative square root, because the domain of our original function only has non-negative numbers.
• Jan 25th 2010, 12:47 PM
emathinstruction
Quote:

Originally Posted by liptonpc
if $f(x)=x^2-9$
$f^{-1}(x)=\pm(x+9)^{\frac{1}2}$
and same thing it not inverse?

You use the vertical line test to check if something is a function. Then, if it is a function, you use the horizontal line test to see if it has an inverse. If it does not pass the horizontal line test, then it isn't an invertible function

The reason if because the (fake) inverse function that you would get would spit out 2 values for (almost) every x value when talking about parabolas, as shown by the $\pm$ in the problem above, and when that happens it cant be a function
• Jan 25th 2010, 12:49 PM
liptonpc
yeah #_# because i usually think $x\ge0$ i forgot they didt say this ^^
thanks ^^
• Jan 25th 2010, 01:01 PM
liptonpc
but if dont ask about inverse or not. Just ask Find $f^{-1}(x)$
$f^{-1}(x)=\pm(x+9)^{\frac{1}2}$
or $f^{-1}(x)=(x+9)^{\frac{1}2}$
???
• Jan 25th 2010, 01:25 PM
Jhevon
Quote:

Originally Posted by liptonpc
but if dont ask about inverse or not. Just ask Find $f^{-1}(x)$
$f^{-1}(x)=\pm(x+9)^{\frac{1}2}$
or $f^{-1}(x)=(x+9)^{\frac{1}2}$
haha, asking to find $f^{-1}$ is asking to find the inverse. if your function is not one-to-one (or onto), either by itself or by restricting its domain, then it does not have an inverse. the question makes no sense.