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Mathematics 10 Online
OpenStudy (anonymous):

I need help, anyone?

OpenStudy (anonymous):

Q. 2,3, 5?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

...

OpenStudy (anonymous):

ohh i c :)

OpenStudy (anonymous):

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

OpenStudy (anonymous):

is this what is required at option a)?? ...I don't get it well enough (I'm not American/British :S )

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

and ur graph is answer to part 5

OpenStudy (anonymous):

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

OpenStudy (anonymous):

do u know how to do Q.3?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

are you there?

OpenStudy (anonymous):

yea

OpenStudy (anonymous):

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

OpenStudy (anonymous):

or f^-1(x)

OpenStudy (anonymous):

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

OpenStudy (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

OpenStudy (anonymous):

lol okay..

OpenStudy (anonymous):

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

OpenStudy (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?

OpenStudy (slaaibak):

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

OpenStudy (anonymous):

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

OpenStudy (slaaibak):

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

OpenStudy (anonymous):

well then @math456 is right !!...

OpenStudy (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

OpenStudy (anonymous):

y=1-x-xy*

OpenStudy (anonymous):

????

OpenStudy (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)

OpenStudy (slaaibak):

with which number do you need help with?

OpenStudy (anonymous):

Q.2 a,b

OpenStudy (anonymous):

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

OpenStudy (slaaibak):

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

OpenStudy (slaaibak):

I mean, flip it over the line

OpenStudy (anonymous):

|dw:1333412317919:dw| like this?

OpenStudy (anonymous):

fliping it will be|dw:1333412361021:dw|

OpenStudy (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

OpenStudy (slaaibak):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

or the inverse?

OpenStudy (slaaibak):

flip f(x) over y=x

OpenStudy (anonymous):

ohh okay!

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (slaaibak):

noooo |dw:1333412696801:dw|

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