1. Invertible functions

i try do it but only first one #_# 2 and 3 i dont know how can do #_#

Thanks.

2. Originally Posted by liptonpc
i try do it but only first one #_# 2 and 3 i dont know how can do #_#

Thanks.
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

3. $\displaystyle f^{-1}(x)=\sqrt\frac {x + 1}{3}$
is it correct ?

4. Originally Posted by liptonpc
$\displaystyle f^{-1}(x)=\sqrt\frac {x + 1}{3}$
is it correct ?
technically you need a +/- in front of that. which would make $\displaystyle f^{-1}$ not a function.

only bijective functions are invetible, and f isn't bijective

5. sorry i dont understand #_#
so the result is ?

6. 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.

7. $\displaystyle f^{-1}(x)=\sqrt\frac {x + 1}{3}$
when i check
$\displaystyle f(f^{-1}(x))=x$

so why it not inverse?

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

so why it not inverse?
because $\displaystyle f^{-1}(x) \ne \sqrt{\frac {x + 1}3}$, rather, $\displaystyle f^{-1} (x) = ~{\color{red} \pm}~ \sqrt {\frac {x + 1}3}$

which means $\displaystyle f^{-1}$ is not a function at all, since that does not pass the vertical line test.

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

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

now on the other hand, if they had restricted the domain and said, $\displaystyle f(x) = x^2 - 9$ 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.

11. Originally Posted by liptonpc
if $\displaystyle f(x)=x^2-9$
$\displaystyle f^{-1}(x)=\pm(x+9)^{\frac{1}2}$
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 $\displaystyle \pm$ in the problem above, and when that happens it cant be a function

12. yeah #_# because i usually think $\displaystyle x\ge0$ i forgot they didt say this ^^
thanks ^^

13. but if dont ask about inverse or not. Just ask Find $\displaystyle f^{-1}(x)$
$\displaystyle f^{-1}(x)=\pm(x+9)^{\frac{1}2}$
or $\displaystyle f^{-1}(x)=(x+9)^{\frac{1}2}$
???

14. Originally Posted by liptonpc
but if dont ask about inverse or not. Just ask Find $\displaystyle f^{-1}(x)$
$\displaystyle f^{-1}(x)=\pm(x+9)^{\frac{1}2}$
or $\displaystyle f^{-1}(x)=(x+9)^{\frac{1}2}$
haha, asking to find $\displaystyle f^{-1}$ 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.