Sketch the graph of the function f(x)=square root X

[Aka]

a) sketch the graph of the function f(x)=square root X
b) on the same set of axes, graph y=f^-1(x), y=-f^-1(-x)

so I did a chart, but i don't know if its right...I don't think its right

f(x)=squareroot x

X Y
0 0
1 1
4 2
9 3

y=f^-1(x)

X Y
0 0
1 -1
4 -2
9 -3

y=-f^-1(-x)

X Y
0 0
-1 1
-4 2
-9 3

and i don't understand the question
if the two graph of y=-f^-1(-x) is drawn from the graph of y=f(x) by a single reflection, what is the equation of the reflection line?

Thanks

 

[Tide]

The graph of the inverse function will be the reflection of the function about the line y = x which amounts to exchanging the values of x and y in your tables and adjusting accordingly for the negative of the inverse.

 

[HallsofIvy]

a) sketch the graph of the function f(x)=square root X
b) on the same set of axes, graph y=f^-1(x), y=-f^-1(-x)

so I did a chart, but i don't know if its right...I don't think its right

f(x)=squareroot x

X Y
0 0
1 1
4 2
9 3

Yes, this is correct.

y=f^-1(x)

X Y
0 0
1 -1
4 -2
9 -3

No, this is y= -f(x), not y= f^-1(x), the inverse function.
To get the inverse function, just swap the x and y columns.
What is the "opposite" of square root?

[/quote]y=-f^-1(-x)

X Y
0 0
-1 1
-4 2
-9 3 [/quote]
see above.

and i don't understand the question
if the two graph of y=-f^-1(-x) is drawn from the graph of y=f(x) by a single reflection, what is the equation of the reflection line?

Thanks

I don't understand "the two graph". Are you sure that you have copied the problem correctly? Draw the graphs as you were told two. Is there some line so that "flipping" the first graph over that line gives the second graph?

 

[Aka]

^ Thanks

Are you sure that you have copied the problem correctly?

sorry, I did not coppy it corrently, it's not supposed to have the two in it :P

so is it like this?

y=f^-1(x)
X Y
0 0
1 1
2 4
3 9

y=f^-1(-x)
X Y
0 0
-1 1
-2 4
-3 9

 

[Moo Of Doom]

^ Thanks


sorry, I did not coppy it corrently, it's not supposed to have the two in it :P

so is it like this?

y=f^-1(x)
X Y
0 0
1 1
2 4
3 9

y=f^-1(-x)
X Y
0 0
-1 1
-2 4
-3 9

Yes, that's better. But, I believe you forgot the negative in front of the f^-1 in the second equation. Shouldn't that be y = -f^-1(-x)?

 

[Aka]

^ yes, it should

so the y values will be negative