# P2 functions

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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..

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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#2

(Original post by

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..

**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..

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#4

(Original post by

yea i hate functions: what does y = / 1/x / look like

**TheWolf**)yea i hate functions: what does y = / 1/x / look like

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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#5

(Original post by

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.

**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.

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#6

**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..

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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#7

**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..

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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#8

(Original post by

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

**mik1a**)f^-1(x) = 1 +/- root(x+4)

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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#9

(Original post by

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)

**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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#10

(Original post by

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

**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

and the domain of f^-1 (x) is x >= -4 ? is that the same as the range of f(x)?

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#11

(Original post by

Just to clarify... and the domain of f^-1 (x) is x >= -4 ? is that the same as the range of f(x)?

**kimoni**)Just to clarify... and the domain of f^-1 (x) is x >= -4 ? is that the same as the range of f(x)?

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(Original post by

yes..it is supposed to be

**lgs98jonee**)yes..it is supposed to be

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#13

(Original post by

So basically, the range of f^-1 is the domain of f, and the domain of f^-1 is the range of f ?

**imasillynarb**)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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#14

**imasillynarb**)

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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#16

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

Thanks

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(Original post by

Heh, it's funny how late on in the year we learn such handy things...

**kimoni**)Heh, it's funny how late on in the year we learn such handy things...

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#18

(Original post by

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

**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

f(x) = 1/x

Domain of f(x):

f(x) < 0

f(x) > 0

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#19

**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

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#20

**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)

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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