SOLUTION: I have to find the inverse of g(x) = the square root of x -4 In order to find the inverse you switch the x and y. x = square root of y - 4 I have to get the y by itself,

Algebra ->  Square-cubic-other-roots -> SOLUTION: I have to find the inverse of g(x) = the square root of x -4 In order to find the inverse you switch the x and y. x = square root of y - 4 I have to get the y by itself,       Log On


   

Question 54929: I have to find the inverse of g(x) = the square root of x -4
In order to find the inverse you switch the x and y.
x = square root of y - 4
I have to get the y by itself, so I squared both sides, leaving me with:
x^2 = y - 4
x^2 + 4 = y
Is this the final answer?
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Yes, you did it correctly, but you should finish the problem by replacing y with g%5E%28-1%29%28x%29 so you have:
g%5E%28-1%29%28x%29+=+x%5E2%2B4
To check you solution, remember that:
g%28g%5E%28-1%29%29%28x%29+=+x and g%5E%28-1%29%28g%28x%29%29+=+x
Let's look at the first check: g%28x%29+=+sqrt%28x-4%29
g%28g%5E%28-1%29%28x%29%29+=+x
g%28x%5E2%2B4%29+=+sqrt%28%28x%5E2%2B4%29-4%29 Simplifying this, you get:
sqrt%28x%5E2%29+=+x This checks out fine.
Now we'll do the second check:
g%5E%28-1%29%28g%28x%29%29+=+x
g%5E%28-1%29%28sqrt%28x-4%29%29+=+sqrt%28%28x%5E2%2B4%29-4%29 Simplifying this, you get:
sqrt%28x%5E2%29+=+x This also checks.