I need help, anyone?

- anonymous

I need help, anyone?

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- anonymous

Q. 2,3, 5?

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- anonymous

do you have a graph yet? ... I can get you one !

- anonymous

|dw:1333410736966:dw| yea its like this!

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- anonymous

...

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- anonymous

ohh i c :)

- anonymous

so for Q.2 we have to graph the inverse, right?

- 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

two lines... f(x)=4x-8 and y=x

- anonymous

umm we have to find inverse first which is x+8/4=y and graphing it gives me |dw:1333411085687:dw|

- anonymous

\[\Large f^{-1}(x) =\frac{1}{f(x)}\] ... so that should be .
.\[\LARGE f^{-1}(x)=\frac{1}{4x-8}\] is it??

- anonymous

and ur graph is answer to part 5

- anonymous

umm our teacher tell us that u switch letters first & then solve for y to get the inverse..

- anonymous

do u know how to do Q.3?

- anonymous

I know it like I wrote it !! O_O ... show me your way then!

- anonymous

are you there?

- anonymous

yea

- anonymous

i already told u inverse seems like x+8/4=y

- anonymous

or f^-1(x)

- anonymous

what does it mean to test the symmetry accross the line y=x?

- 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

lol okay..

- anonymous

y=f(x)=4x-8
x=4y-8 (switched)
x+8/4=y (solve for y)

- 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

the inverse of f(x) is not 1/f(x)

- anonymous

ok I guess I'm wrong, but I want to learn it.. :(

- slaaibak

To find the inverse, switch x and y, and write in terms of y.

- anonymous

well then @math456 is right !!...

- 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

y=1-x-xy*

- anonymous

????

- 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

with which number do you need help with?

- anonymous

Q.2 a,b

- anonymous

i mean like for a we hv to reflect the graph of f across the diagonal line y=x, how?

- slaaibak

draw the y=x line and flip the line over it

- slaaibak

I mean, flip it over the line

- anonymous

|dw:1333412317919:dw| like this?

- anonymous

fliping it will be|dw:1333412361021:dw|

- 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

Noo... you flip the function f(x) over the line y=x.

- anonymous

so for Q2a we hv to flip over the f(x) ?

- anonymous

or the inverse?

- slaaibak

flip f(x) over y=x

- anonymous

ohh okay!

- anonymous

|dw:1333412477378:dw| so its gona be like this?

- anonymous

how about 2b? and question 5 ask the same thing as 2a, right?

- slaaibak

noooo
|dw:1333412696801:dw|

- slaaibak

question 2 b: get a few points (x,y)
then change them to (y,x) and draw them

- anonymous

and 5 is |dw:1333412939248:dw| ?

- slaaibak

Not exactly sure what they mean..

- anonymous

so for 2b i got a graph like this |dw:1333413313588:dw|

- 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

that was for f(x) and if I reflect it, i'll get |dw:1333413484096:dw|

- anonymous

with Question 5?

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