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

Question

anonymous · Answer

@Hero

anonymous · Answer

@Astrophysics

anonymous · Answer

@UsukiDoll

campbell_st · Answer

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 · Answer

So should i solve for x? @campbell_st

campbell_st · Answer

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

anonymous · Answer

y=4− x√ 2,4+ x√ 2

campbell_st · Answer

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

campbell_st · Answer

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

anonymous · Answer

@UsukiDoll

anonymous · Answer

It doesnt match any of my answers

campbell_st · Answer

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

anonymous · Answer

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

anonymous · Answer

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

usukidoll · Answer

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

anonymous · Answer

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

usukidoll · Answer

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

anonymous · Answer

haha I just want to make sure the answer is correct

usukidoll · Answer

$$x = (8 - 2y)^2$$ [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 · Answer

subtracr

usukidoll · Answer

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

anonymous · Answer

Add 2?

usukidoll · Answer

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

anonymous · Answer

divide?

usukidoll · Answer

yes we divide -2 on both sides.

usukidoll · Answer

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

usukidoll · Answer

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

anonymous · Answer

So his answer was right than

usukidoll · Answer

so what's -8/-2 ?

anonymous · Answer

-4

usukidoll · Answer

a negative divided by a negative turns positive

anonymous · Answer

4

usukidoll · Answer

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

usukidoll · Answer

$$\frac{-\sqrt{x}}{2}+4$$

anonymous · Answer

ok this is my final question tonight lol

anonymous · Answer

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

anonymous · Answer

do we put 4 in all the x's?

usukidoll · Answer

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

usukidoll · Answer

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

anonymous · Answer

solve for y?

usukidoll · Answer

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

anonymous · Answer

the x and y

usukidoll · Answer

yes
now we solve for y 
$$x=y^2-12$$

usukidoll · Answer

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

anonymous · Answer

add

usukidoll · Answer

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

anonymous · Answer

divide?

usukidoll · Answer

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

anonymous · Answer

nothing?

usukidoll · Answer

$$\sqrt{y^2} = ? $$

usukidoll · Answer

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

usukidoll · Answer

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

anonymous · Answer

1

anonymous · Answer

it would just turn into y

usukidoll · Answer

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 · Answer

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

usukidoll · Answer

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

usukidoll · Answer

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

anonymous · Answer

haha ok

usukidoll · Answer

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 · Answer

Is the answer 6?

usukidoll · Answer

no try placing f inverse in f(x)

anonymous · Answer

4?

usukidoll · Answer

yeah

anonymous · Answer

sweetness

anonymous · Answer

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

usukidoll · Answer

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

anonymous · Answer

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

anonymous · Answer

@UsukiDoll

usukidoll · Answer

oh that's awesome! :D