The function $f$ is defined by $$f(x) = \frac{1}{2}(x^2 + 1), \, x \geq 0$$ 6 (a) Find the range of $f$ - AQA - A-Level Maths Mechanics - Question 6 - 2019 - Paper 1

To find the inverse function, we start with:

$y = \frac{1}{2}(x^2 + 1)$

Rearranging gives:

1. Multiply by 2: $2y = x^2 + 1$
2. Subtract 1: $x^2 = 2y - 1$
3. Take the square root: $x = \sqrt{2y - 1}$
4. Thus: $f^{-1}(y) = \sqrt{2y - 1}, \text{ for } y \geq \frac{1}{2}$

The range of the inverse function $f^{-1}(x)$ corresponds to the domain of the original function $f(x)$.

Since $f(x)$ is defined for $x \geq 0$, the range of $f^{-1}(x)$ is: $x \geq 0$