Solve for the inverse of f(x)=−2x^2−1

Question

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

Answer 1

To find the inverse of the function \( f(x) = -2x^2 - 1 \) with the restriction that the domain is \( x \geq 0 \), we can follow these steps:

1. Replace \( f(x) \) with \( y \): \[ y = -2x^2 - 1 \]

2. Solve for \( x \) in terms of \( y \): First, isolate the \( x^2 \) term: \[ y + 1 = -2x^2 \] \[ -\frac{y + 1}{2} = x^2 \] Now we can express \( x \) by taking the square root of both sides. Since we are restricted to \( x \geq 0 \), we take the positive root: \[ x = \sqrt{-\frac{y + 1}{2}} \]

3. Express the inverse function: Swap \( x \) and \( y \) to find the inverse function: \[ f^{-1}(x) = \sqrt{-\frac{x + 1}{2}} \]

Thus, the inverse function is: \[ f^{-1}(x) = \sqrt{-\frac{x + 1}{2}} \]

Note:

The domain of the function \( f(x) \) for \( x \geq 0 \) leads to a range of \( y \) values such that \( y \leq -1 \). Therefore, the domain of the inverse function \( f^{-1}(x) \) is \( x \leq -1 \).