anonymous
  • anonymous
I need help, anyone?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Q. 2,3, 5?
anonymous
  • anonymous
do you have a graph yet? ... I can get you one !
anonymous
  • anonymous
|dw:1333410736966:dw| yea its like this!

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anonymous
  • anonymous
...
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anonymous
  • anonymous
ohh i c :)
anonymous
  • anonymous
so for Q.2 we have to graph the inverse, right?
anonymous
  • anonymous
is this what is required at option a)?? ...I don't get it well enough (I'm not American/British :S )
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anonymous
  • anonymous
two lines... f(x)=4x-8 and y=x
anonymous
  • anonymous
umm we have to find inverse first which is x+8/4=y and graphing it gives me |dw:1333411085687:dw|
anonymous
  • anonymous
\[\Large f^{-1}(x) =\frac{1}{f(x)}\] ... so that should be . .\[\LARGE f^{-1}(x)=\frac{1}{4x-8}\] is it??
anonymous
  • anonymous
and ur graph is answer to part 5
anonymous
  • anonymous
umm our teacher tell us that u switch letters first & then solve for y to get the inverse..
anonymous
  • anonymous
do u know how to do Q.3?
anonymous
  • anonymous
I know it like I wrote it !! O_O ... show me your way then!
anonymous
  • anonymous
are you there?
anonymous
  • anonymous
yea
anonymous
  • anonymous
i already told u inverse seems like x+8/4=y
anonymous
  • anonymous
or f^-1(x)
anonymous
  • anonymous
what does it mean to test the symmetry accross the line y=x?
anonymous
  • anonymous
I don't know it either... but tell me, how do you get x+8/4=y ... show me steps I want to see them if it's possible ! O_O
anonymous
  • anonymous
lol okay..
anonymous
  • anonymous
y=f(x)=4x-8 x=4y-8 (switched) x+8/4=y (solve for y)
anonymous
  • anonymous
still dizzy... I don't know.. OK now tell me if we have \[\LARGE f(x)=\frac{1-x}{1+x}\] how does f^{-1} go?
slaaibak
  • slaaibak
the inverse of f(x) is not 1/f(x)
anonymous
  • anonymous
ok I guess I'm wrong, but I want to learn it.. :(
slaaibak
  • slaaibak
To find the inverse, switch x and y, and write in terms of y.
anonymous
  • anonymous
well then @math456 is right !!...
anonymous
  • anonymous
f(x)=1-x/1+x x=1-y/1+y x(1+y)=1-y x+xy=1-y y=1-x+xy
anonymous
  • anonymous
y=1-x-xy*
anonymous
  • anonymous
????
slaaibak
  • slaaibak
y = (1 - x)/(1 + x) Switching the x and y: x = (1-y)/(1+y) x + xy = 1 - y y(1+x) = 1-x y = (1-x)/(x+1)
slaaibak
  • slaaibak
with which number do you need help with?
anonymous
  • anonymous
Q.2 a,b
anonymous
  • anonymous
i mean like for a we hv to reflect the graph of f across the diagonal line y=x, how?
slaaibak
  • slaaibak
draw the y=x line and flip the line over it
slaaibak
  • slaaibak
I mean, flip it over the line
anonymous
  • anonymous
|dw:1333412317919:dw| like this?
anonymous
  • anonymous
fliping it will be|dw:1333412361021:dw|
slaaibak
  • slaaibak
That's y=x yes. Best way to see it is to draw the line, then folding the paper over y=x and to see how the graph reflects
slaaibak
  • slaaibak
Noo... you flip the function f(x) over the line y=x.
anonymous
  • anonymous
so for Q2a we hv to flip over the f(x) ?
anonymous
  • anonymous
or the inverse?
slaaibak
  • slaaibak
flip f(x) over y=x
anonymous
  • anonymous
ohh okay!
anonymous
  • anonymous
|dw:1333412477378:dw| so its gona be like this?
anonymous
  • anonymous
how about 2b? and question 5 ask the same thing as 2a, right?
slaaibak
  • slaaibak
noooo |dw:1333412696801:dw|
slaaibak
  • slaaibak
question 2 b: get a few points (x,y) then change them to (y,x) and draw them
anonymous
  • anonymous
and 5 is |dw:1333412939248:dw| ?
slaaibak
  • slaaibak
Not exactly sure what they mean..
anonymous
  • anonymous
so for 2b i got a graph like this |dw:1333413313588:dw|
slaaibak
  • slaaibak
... I honestly don't know how you got that. Look at my graph. Thats how f, y=x and f^-1 looks like
anonymous
  • anonymous
that was for f(x) and if I reflect it, i'll get |dw:1333413484096:dw|
anonymous
  • anonymous
with Question 5?

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