How to find the inverse function?

We are given the function y = square root of (x² - 6x + 15), the domain of the function is x <= 3 and x > negative infinity, The question ask for the inverse function f^-1(x). Now we express the function in maximum form.

y = square root of (x² - 6x + 15)

= square root of (x² - 6x + 3² - 3² + 15)

= square root of (x² - 6x + 3² - 9 + 15)

= square root of (x² - 6x + 3² + 6)

= square root of [(x - 3)² + 6]

Now we will find the range of y from the domain of the function. When x tend to negative infinity, y tend to positive infinity. When x = 3, y equals to the square root of 6. So the range of y is y >= square root of 6 and y < positive infinity.

Now we are going to find the inverse function f^-1(x).

y = square root of [(x - 3)² + 6]

y² = (x - 3)² + 6

now express x in terms of y

(x - 3)² = y² - 6

x - 3 = either positive or negative of square root of (y² - 6)

since x is <= 3 (given), so (x - 3) <= 0

so x = 3 = - square root of (y² - 6) and negative infinity < y <= square root of 6.

now change y to x and change x to y

y = 3 - square root of (x² - 6) and negative infinity < x <= square root of 6.

so we get the inverse function

f^-1(x) = 3 - square root of (x² - 6) and negative infinity < x <= square root of 6.