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1. 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..
2. (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
3. yea i hate functions: what does y = / 1/x / look like
4. (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.
5. (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!
6. (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)
7. (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
8. (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
9. (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?
10. (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)?
11. (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
12. (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 ?
13. (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...
14. (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
15. (Original post by TheWolf)
yes
aaah! that's useful!
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
17. (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
18. (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
19. (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
20. (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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