For the function f(x) = (8-2x)2 ,find f-1 . Determine whether f-1 is a function.

- anonymous

For the function f(x) = (8-2x)2 ,find f-1 . Determine whether f-1 is a function.

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- anonymous

@Hero

- anonymous

@Astrophysics

- anonymous

@UsukiDoll

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## More answers

- campbell_st

just start by swapping x and y so the original equation is
\[y = (8 - 2x)^2\]
do the swap
\[x = (8 - 2y)^2\]
now may y the subject of the equation

- anonymous

So should i solve for x? @campbell_st

- campbell_st

no you should solve for y...
1st step is to take the square root of both sides of the equation...

- anonymous

y=4− x√ 2,4+ x√ 2

- campbell_st

well the problem is that the original equation is a function, the inverse will be a relationship... and your solution seems ok...

- campbell_st

the way you test the inverse is a function is to use the vertical line test after graphing the curve

- anonymous

@UsukiDoll

- anonymous

It doesnt match any of my answers

- campbell_st

so do you have
\[f^{-1}(x) = \frac{8 - \sqrt{x}}{2}\]

- anonymous

idk i cant see my answer A lol it's probably that one than

- anonymous

All my other answers are a + not a - and I cant see answer A

- UsukiDoll

take the one with the + ? Idk.. but @campbell_st is right. you switch the signs of x and y and solve for y

- anonymous

Is the final answer correct? If so I'll just chose A because everything else is a + after the 8

- UsukiDoll

?!!!! ok looks like I have to square root both sides or I won't go further lol

- anonymous

haha I just want to make sure the answer is correct

- UsukiDoll

\[x = (8 - 2y)^2\] [square both sides so we have y by itself]
\[\sqrt{x}=8-2y\]
ok so I'm just gonna stop at this point. I don't want to get busted :P
but anyway we need y by itself so I ____________ 8 from both sides

- anonymous

subtracr

- UsukiDoll

yes \[\sqrt{x}-8=-2y\] so now what do we need to do to get y by itself?

- anonymous

Add 2?

- UsukiDoll

not quite. We can't use addition... try again

- anonymous

divide?

- UsukiDoll

yes we divide -2 on both sides.

- UsukiDoll

\[\frac{\sqrt{x}-8}{-2}=y\] we can simplify further

- UsukiDoll

for starters we can split this equation up....
\[\frac{\sqrt{x}-8}{-2}=y \rightarrow \frac{-\sqrt{x}}{2}+\frac{-8}{-2}\]

- anonymous

So his answer was right than

- UsukiDoll

so what's -8/-2 ?

- anonymous

-4

- UsukiDoll

a negative divided by a negative turns positive

- anonymous

4

- UsukiDoll

yeah.. and about his answer..I think he dropped the negative sign

- UsukiDoll

\[\frac{-\sqrt{x}}{2}+4\]

- anonymous

ok this is my final question tonight lol

- anonymous

For the function f(x) = x2 - 12, find (f o f-1)(4)

- anonymous

do we put 4 in all the x's?

- UsukiDoll

hmmmm looks like we need to find f^{-1} first and plug that guy into f(x)

- UsukiDoll

\[f(x) =x^2-12\] f(x) is also y
\[y=x^2-12 \]
now how do we find the inverse ?

- anonymous

solve for y?

- UsukiDoll

that's the second step
what is the first step? what needs to be switched?

- anonymous

the x and y

- UsukiDoll

yes
now we solve for y
\[x=y^2-12\]

- UsukiDoll

so how do we get y by itself?
we need to __________ 12 to both sides

- anonymous

add

- UsukiDoll

\[x+12=y^2 \]
good
now we need to ___________ both sides
we need a y by itself... it can't have an exponent

- anonymous

divide?

- UsukiDoll

no. umm.. what is the square root of y^2 ?

- anonymous

nothing?

- UsukiDoll

\[\sqrt{y^2} = ? \]

- UsukiDoll

it can't be nothing.. otherwise we won't have y ._.

- UsukiDoll

maybe exponential form will be easier \[\large y^\frac{2}{2}\] what's 2/2 ?

- anonymous

1

- anonymous

it would just turn into y

- UsukiDoll

yes but keep in mind since we took the square root of y^2 we have to take the square root of the left hand side as well
\[\sqrt{x+12}=y\]

- UsukiDoll

eep \[f^{-1}(x) = \sqrt{x+12} \]

- UsukiDoll

now we will place that inverse function into our \[f(x) = x^2-12 \]
but something unusual is going to happen

- UsukiDoll

oh wait it won't sorry
we will still have something ok good whew

- anonymous

haha ok

- UsukiDoll

so f o f^{-1} means place f^{-1} inside the f(x) function.. yeah I was just looking at the number and not the variable xD

- anonymous

Is the answer 6?

- UsukiDoll

no try placing f inverse in f(x)

- anonymous

4?

- UsukiDoll

yeah

- anonymous

sweetness

- anonymous

im gonna need for you to be online whenever i am online lol

- UsukiDoll

\[f^{-1}(x) = \sqrt{x+12} , f(x) = x^2-12 \]
\[f(x) \cdot f^{-1}(x) \]
\[f^{-1}(x) \cdot f(x) = (\sqrt{x+12})^2-12 \]
\[f^{-1}(x) \cdot f(x) = x+12-12 \]
\[f^{-1}(x) \cdot f(x) =x \]
when x = 4
\[f^{-1}(x) \cdot f(x) =4 \]

- anonymous

Thank you for all of your help, I got an A on my hw

- anonymous

@UsukiDoll

- UsukiDoll

oh that's awesome! :D

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