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

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campbell_st
  • 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
  • anonymous
So should i solve for x? @campbell_st
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
  • anonymous
y=4− x√ 2,4+ x√ 2
campbell_st
  • 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
  • campbell_st
the way you test the inverse is a function is to use the vertical line test after graphing the curve
anonymous
  • anonymous
anonymous
  • anonymous
It doesnt match any of my answers
campbell_st
  • campbell_st
so do you have \[f^{-1}(x) = \frac{8 - \sqrt{x}}{2}\]
anonymous
  • anonymous
idk i cant see my answer A lol it's probably that one than
anonymous
  • anonymous
All my other answers are a + not a - and I cant see answer A
UsukiDoll
  • 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
  • anonymous
Is the final answer correct? If so I'll just chose A because everything else is a + after the 8
UsukiDoll
  • UsukiDoll
?!!!! ok looks like I have to square root both sides or I won't go further lol
anonymous
  • anonymous
haha I just want to make sure the answer is correct
UsukiDoll
  • 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
  • anonymous
subtracr
UsukiDoll
  • UsukiDoll
yes \[\sqrt{x}-8=-2y\] so now what do we need to do to get y by itself?
anonymous
  • anonymous
Add 2?
UsukiDoll
  • UsukiDoll
not quite. We can't use addition... try again
anonymous
  • anonymous
divide?
UsukiDoll
  • UsukiDoll
yes we divide -2 on both sides.
UsukiDoll
  • UsukiDoll
\[\frac{\sqrt{x}-8}{-2}=y\] we can simplify further
UsukiDoll
  • UsukiDoll
for starters we can split this equation up.... \[\frac{\sqrt{x}-8}{-2}=y \rightarrow \frac{-\sqrt{x}}{2}+\frac{-8}{-2}\]
anonymous
  • anonymous
So his answer was right than
UsukiDoll
  • UsukiDoll
so what's -8/-2 ?
anonymous
  • anonymous
-4
UsukiDoll
  • UsukiDoll
a negative divided by a negative turns positive
anonymous
  • anonymous
4
UsukiDoll
  • UsukiDoll
yeah.. and about his answer..I think he dropped the negative sign
UsukiDoll
  • UsukiDoll
\[\frac{-\sqrt{x}}{2}+4\]
anonymous
  • anonymous
ok this is my final question tonight lol
anonymous
  • anonymous
For the function f(x) = x2 - 12, find (f o f-1)(4)
anonymous
  • anonymous
do we put 4 in all the x's?
UsukiDoll
  • UsukiDoll
hmmmm looks like we need to find f^{-1} first and plug that guy into f(x)
UsukiDoll
  • UsukiDoll
\[f(x) =x^2-12\] f(x) is also y \[y=x^2-12 \] now how do we find the inverse ?
anonymous
  • anonymous
solve for y?
UsukiDoll
  • UsukiDoll
that's the second step what is the first step? what needs to be switched?
anonymous
  • anonymous
the x and y
UsukiDoll
  • UsukiDoll
yes now we solve for y \[x=y^2-12\]
UsukiDoll
  • UsukiDoll
so how do we get y by itself? we need to __________ 12 to both sides
anonymous
  • anonymous
add
UsukiDoll
  • 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
  • anonymous
divide?
UsukiDoll
  • UsukiDoll
no. umm.. what is the square root of y^2 ?
anonymous
  • anonymous
nothing?
UsukiDoll
  • UsukiDoll
\[\sqrt{y^2} = ? \]
UsukiDoll
  • UsukiDoll
it can't be nothing.. otherwise we won't have y ._.
UsukiDoll
  • UsukiDoll
maybe exponential form will be easier \[\large y^\frac{2}{2}\] what's 2/2 ?
anonymous
  • anonymous
1
anonymous
  • anonymous
it would just turn into y
UsukiDoll
  • 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
  • UsukiDoll
eep \[f^{-1}(x) = \sqrt{x+12} \]
UsukiDoll
  • UsukiDoll
now we will place that inverse function into our \[f(x) = x^2-12 \] but something unusual is going to happen
UsukiDoll
  • UsukiDoll
oh wait it won't sorry we will still have something ok good whew
anonymous
  • anonymous
haha ok
UsukiDoll
  • 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
  • anonymous
Is the answer 6?
UsukiDoll
  • UsukiDoll
no try placing f inverse in f(x)
anonymous
  • anonymous
4?
UsukiDoll
  • UsukiDoll
yeah
anonymous
  • anonymous
sweetness
anonymous
  • anonymous
im gonna need for you to be online whenever i am online lol
UsukiDoll
  • 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
  • anonymous
Thank you for all of your help, I got an A on my hw
anonymous
  • anonymous
UsukiDoll
  • UsukiDoll
oh that's awesome! :D

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