Find the inverse function of f(x)= (√9x-5) +2 for x>/ 5/9

Answer 1

Your parentheses are a bit confusing but I'm going to assume the function is actually f(x) = √(9x - 5) + 2

1. Write the equation as y = √(9x - 5) + 2

2. Swap "x" and "y" to make it x = √(9y - 5) + 2

3. Solve for y:

x = √(9y - 5) + 2

x - 2 = √(9y - 5)

(x - 2)² = 9y - 5

(x - 2)² + 5 = 9y

1/9(x - 2)² + 5/9 = y

4. Re-write the inverse as f^-1(x) by replacing "y" with f^-1(x):

f^-1(x) = 1/9(x - 2)² + 5/9 Since there is a domain restriction on the original function, we must apply the same restriction here but the inverse switches the domain and range so we must know the range of the original function. So f(5/9) = √(9•(5/9) - 5) + 2 = 2. So the range is y > 2. That makes the domain of the inverse x > 2.

So the final answer would be f^-1(x) = 1/9(x - 2)² + 5/9 for x > 2.