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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

- anonymous

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

- jamiebookeater

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

- anonymous

- anonymous

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## 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

- 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

oh that's awesome! :D

Looking for something else?

Not the answer you are looking for? Search for more explanations.