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Barny
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x^2 - 2x - 3, x is real, x >/= 1

Write down the domain and range of f^-1

Can someone explain this to me? I dont get this function stuff at all..
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Silly Sally
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(Original post by imasillynarb)
x^2 - 2x - 3, x is real, x >/= 1

Write down the domain and range of f^-1

Can someone explain this to me? I dont get this function stuff at all..
I think probably the best thing is to draw the graph and work out the lowest/highest point to determine the range. Don't know if that helps - i have a problem with functions as well
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TheWolf
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yea i hate functions: what does y = / 1/x / look like
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john !!
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(Original post by TheWolf)
yea i hate functions: what does y = / 1/x / look like
If you mean y = | 1/x |, then it is just reflected on the x-axis below x=0, because y must always be positive.

However, if you have y = 1/|x|, then below x=0 it is reflected on the y-axis, because the graph acts as if x is positive all the time.
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TheWolf
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(Original post by mik1a)
If you mean y = | 1/x |, then it is just reflected on the x-axis below x=0, because y must always be positive.

However, if you have y = 1/|x|, then below x=0 it is reflected on the y-axis, because the graph acts as if x is positive all the time.
thanks!
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john !!
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(Original post by imasillynarb)
x^2 - 2x - 3, x is real, x >/= 1

Write down the domain and range of f^-1

Can someone explain this to me? I dont get this function stuff at all..
f(x) = x^2 - 2x - 3

y = f(x)

Change round:

x = f(y)

so if:

y = x^2 - 2x - 3

We change to:

x = y^2 - 2y - 3
x = (y - 1)^2 - 4
x + 4 = (y - 1)^2
+/- root(x+4) = y - 1
y = 1 +/- root(x+4)

Therefore:

f^-1(x) = 1 +/- root(x+4)
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john !!
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(Original post by imasillynarb)
x^2 - 2x - 3, x is real, x >/= 1

Write down the domain and range of f^-1

Can someone explain this to me? I dont get this function stuff at all..
Because this is the inverse of the function f, the ranges and domains will swap round.

This works because the function is one-one mapping, so only one f(x) value can be found from one x value.

So the range is clearly x >_ 1.
The domain:

y = x^2 - 2x - 3
y = (x - 1)^2 - 4

Showing a min. point of -4.
The domain is therefore:

f(x) >_ -4
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lgs98jonee
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(Original post by mik1a)

f^-1(x) = 1 +/- root(x+4)
u ignore the -ve root though as functions and their inverses must be 1 to 1 i think. therefore, you can not have two values of x that make one of y nor two ys that give an x.
so f^-1(x)=1+ l rt(x+4) l

domain of inverse is the same as range of normal
so the range of this will be f^-1(x) >/= 1

and domain is x>=-4
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TheWolf
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(Original post by mik1a)
f(x) = x^2 - 2x - 3

y = f(x)

Change round:

x = f(y)

so if:

y = x^2 - 2x - 3

We change to:

x = y^2 - 2y - 3
x = (y - 1)^2 - 4
x + 4 = (y - 1)^2
+/- root(x+4) = y - 1
y = 1 +/- root(x+4)

Therefore:

f^-1(x) = 1 +/- root(x+4)


mmmm i drew the graph out, and i know that x is bigger or equal to one. when x = 1, y = -4, and it is also the minimum point. so isnt the domain > or equal to -4?
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kimoni
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(Original post by mik1a)
Because this is the inverse of the function f, the ranges and domains will swap round.

This works because the function is one-one mapping, so only one f(x) value can be found from one x value.

So the range is clearly x >_ 1.
The domain:

y = x^2 - 2x - 3
y = (x - 1)^2 - 4

Showing a min. point of -4.
The domain is therefore:

f(x) >_ -4
Just to clarify... the domain of f(x) is x >= 1
and the domain of f^-1 (x) is x >= -4 ? is that the same as the range of f(x)?
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lgs98jonee
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(Original post by kimoni)
Just to clarify... and the domain of f^-1 (x) is x >= -4 ? is that the same as the range of f(x)?
yes..it is supposed to be
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Barny
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(Original post by lgs98jonee)
yes..it is supposed to be
So basically, the range of f^-1 is the domain of f, and the domain of f^-1 is the range of f ?
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kimoni
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(Original post by imasillynarb)
So basically, the range of f^-1 is the domain of f, and the domain of f^-1 is the range of f ?
Heh, it's funny how late on in the year we learn such handy things...
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TheWolf
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(Original post by imasillynarb)
So basically, the range of f^-1 is the domain of f, and the domain of f^-1 is the range of f ?
yes
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kimoni
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(Original post by TheWolf)
yes
aaah! that's useful!
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Silly Sally
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So when asked to find a range of a function we find its max/min point and say the range is higher than or lower than a specific point? Also, if you were trying to the domain of a function and there was one point where x was an asymptote, how would you express the domain???

Thanks
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Barny
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(Original post by kimoni)
Heh, it's funny how late on in the year we learn such handy things...
Yeh...wish my teacher would of said that
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john !!
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(Original post by Silly Sally)
So when asked to find a range of a function we find its max/min point and say the range is higher than or lower than a specific point? Also, if you were trying to the domain of a function and there was one point where x was an asymptote, how would you express the domain???

Thanks
eg.

f(x) = 1/x

Domain of f(x):

f(x) < 0
f(x) > 0
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TheWolf
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(Original post by Silly Sally)
So when asked to find a range of a function we find its max/min point and say the range is higher than or lower than a specific point? Also, if you were trying to the domain of a function and there was one point where x was an asymptote, how would you express the domain???

Thanks
to find range of function, because range is to do with the y values, just draw the graph, and find the x value of minimum or maximum point, and put this x value into the curve's equation to find y value of the minimum/maximum point
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hihihihi
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(Original post by mik1a)
f(x) = x^2 - 2x - 3

y = f(x)

Change round:

x = f(y)

so if:

y = x^2 - 2x - 3

We change to:

x = y^2 - 2y - 3
x = (y - 1)^2 - 4
x + 4 = (y - 1)^2
+/- root(x+4) = y - 1
y = 1 +/- root(x+4)

Therefore:

f^-1(x) = 1 +/- root(x+4)
yes so that means the domain of f^-1(x) > -4 ?



you could also do it a different way: just complete the square to get
0=(x-1)^2 -4


so minima of f(x) is (1,-4) so the range of f(x)>-4,
therefore domain of f^-1(x)>-4
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