Question

How do you find the inverse of f(x)= 1+ square root of 1-x2

Answer 1

\[x=1+\sqrt{1-y^2}\]

Solve for y.

Answer 2

the equation is \[f(x)=1+\sqrt{1-x ^{2}}\]

Answer 3

hupeorekef

Answer 4

Yes but in inverse, the x and y switch

Answer 5

ookay...

Answer 6

\[x^{2}-2x=-y ^{2}\]

Answer 7

this is what i got so far.

Answer 8

how do you solve for y?

Answer 9

How -2x?

Answer 10

\[(x-1)^{2}\]

Answer 11

Ok, that should be fine then. Now make -y^2 positive by multiplying both sides by -1 so you can take the square root.

Answer 12

And remember that square-roots have 2 solutions: one positive and one negative.

Answer 13

\[y=\pm \sqrt{-x ^{2}+2x}\]

Answer 14

is that right?

Answer 15

Should all look like this up to that point:
Starting with
\[x=1+\sqrt{1-y^2}\]
Subtract 1 from both sides, then square both sides
\[(x-1)^2=1-y^2\]
Subtract 1 from both sides, then multiply both sides by -1 (Could also multiply everything by -1 first and then add 1 to both sides..)
\[y^2=-(x-1)^2+1 \rightarrow y=±\sqrt{-(x-1)^2+1}\]
You can then simplify the radicand if you like.

Answer 16

Yep, looks like you got it.

Answer 17

how come when i plug in like -1 or 0 in there the equation is wrong?

Answer 18

What do you mean by "wrong?"

Answer 19

if you plug in -1 for x you will get a negative answer in the square root...

Answer 20

Then -1 is not in the domain of x over the real numbers.

Answer 21

in this question they gave me the range from \[-1\le(x) \le0\]

Answer 22

In your original function, any x >1 is outside the domain. Establishing the domain of the function and its inverse is part of the process.

Answer 23

how would i graph the inverse?

Answer 24

The original function is the top half of a circle centered at (0,1) with a radius of 1.
The inverse is the right half of a circle centered at (1,0) with a radius of 1.

Answer 25

okay thx