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

To find the inverse function $f^{-1}(x)$, we start from the definition of $y$ in terms of $x$:

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

Rearranging gives:

1. Multiply both sides by 2: $2y = x^2 + 1$

2. Subtract 1 from both sides: $x^2 = 2y - 1$

3. Take the square root: $x = \sqrt{2y - 1}$

Since $x \geq 0$, we have the inverse function:

$f^{-1}(y) = \sqrt{2y - 1}, \quad y \geq \frac{1}{2}$

To determine the range of the inverse function $f^{-1}(x) = \sqrt{2x - 1}$, we analyze its behavior:

Since $x \geq \frac{1}{2}$ (from the range of $f$), the smallest value of $f^{-1}(x)$ occurs at $x = \frac{1}{2}$: $f^{-1}\left(\frac{1}{2}\right) = \sqrt{2 \cdot \frac{1}{2} - 1} = 0$

As $x$ increases beyond rac{1}{2}, $f^{-1}(x)$ increases without bound: $\lim_{x \to \infty} f^{-1}(x) = \infty$

Thus, the range of $f^{-1}(x)$ is: $\{y | y \geq 0\}$