Mathematics
10 Online
OpenStudy (anonymous):
I need help, anyone?
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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 :)
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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??
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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?
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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
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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.. :(
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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):
????
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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
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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.
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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?
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OpenStudy (anonymous):
how about 2b? and question 5 ask the same thing as 2a, right?
OpenStudy (slaaibak):
noooo
|dw:1333412696801:dw|