Find the inverse of the translated quadratic function y=x^2+11

with domain restricted to x≥0
.(1 point)

1 answer

To find the inverse of the function \( y = x^2 + 11 \) with the domain restricted to \( x \geq 0 \), we can follow these steps:

  1. Express \( x \) in terms of \( y \): Start with the original equation: \[ y = x^2 + 11. \]

  2. Subtract 11 from both sides: \[ y - 11 = x^2. \]

  3. Take the square root of both sides: Since we are only considering the domain where \( x \geq 0 \), we take the positive square root: \[ x = \sqrt{y - 11}. \]

  4. Solve for \( y \): Now we express \( y \) in terms of \( x \): \[ y = x^2 + 11. \]

    To find the inverse, we swap \( x \) and \( y \): \[ x = \sqrt{y - 11}. \]

  5. Solve for \( y \) in terms of \( x \): Rearranging the expression gives: \[ y - 11 = x^2. \] \[ y = x^2 + 11. \]

Since we need the inverse function, we write it as: \[ y = x^2 - 11. \]

Thus, the inverse function is: \[ f^{-1}(x) = \sqrt{x - 11}. \]

  1. Specifying the domain: Since we originally had \( x^2 + 11 \) and restricted \( x \) to \( x \geq 0 \), the range of the original function is \( [11, \infty) \). Therefore, the domain of the inverse function will also be \( [11, \infty) \).

In conclusion, the inverse of the translated quadratic function \( y = x^2 + 11 \) for \( x \geq 0 \) is: \[ f^{-1}(x) = \sqrt{x - 11}, \quad \text{for } x \geq 11. \]