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.

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- Jan 24th 2010, 08:42 AMliptonpcInvertible 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 AMJhevon
remember, $\displaystyle f^{-1}$ undoes $\displaystyle f$. This means, the output becomes the input. Let $\displaystyle y = f^{-1}(x)$, then

$\displaystyle x = 3y^2 - 1$

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

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

$

is it correct ? - Jan 24th 2010, 12:31 PMJhevon
- Jan 24th 2010, 01:22 PMliptonpc
sorry i dont understand #_#

so the result is ? - Jan 24th 2010, 02:16 PMJhevon
- Jan 25th 2010, 12:26 PMliptonpc
$\displaystyle f^{-1}(x)=\sqrt\frac {x + 1}{3}$

when i check

$\displaystyle f(f^{-1}(x))=x$

so why it not inverse? - Jan 25th 2010, 12:31 PMJhevon
- Jan 25th 2010, 12:41 PMliptonpc
if $\displaystyle f(x)=x^2-9$

$\displaystyle f^{-1}(x)=\pm(x+9)^{\frac{1}2}$

and same thing it not inverse? - Jan 25th 2010, 12:45 PMJhevon
correct.

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

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 $\displaystyle \pm$ in the problem above, and when that happens it cant be a function - Jan 25th 2010, 12:49 PMliptonpc
yeah #_# because i usually think $\displaystyle x\ge0$ i forgot they didt say this ^^

thanks ^^ - Jan 25th 2010, 01:01 PMliptonpc
but if dont ask about inverse or not. Just ask Find $\displaystyle f^{-1}(x)$

so i have to answer

$\displaystyle f^{-1}(x)=\pm(x+9)^{\frac{1}2}$

or $\displaystyle f^{-1}(x)=(x+9)^{\frac{1}2}$

??? - Jan 25th 2010, 01:25 PMJhevon
- Jan 25th 2010, 01:30 PMliptonpc
^^ thanks u ^^