Let g(x) = x - 6√x + 1, 0 ≤ x ≤ 9. Find g⁻¹(x) and verify (g ∘ g⁻¹)(4) = 4.

Let $g(x) = x - 6\sqrt{x} + 1, 0 \le x \le 9$. Find $g^{-1}(x)$ and verify $(g \circ g^{-1})(4) = 4$.

Answer

Official answer key

On $0 \le x \le 9$, $\sqrt{x} \le 3$, so $\sqrt{x} = 3 - \sqrt{y+8}$ and $g^{-1}(x) = \left(3 - \sqrt{x+8}\right)^2$. Verification: $g\left(g^{-1}(x)\right) = \left(3 - \sqrt{x+8}\right)^2 - 6\left(3 - \sqrt{x+8}\right) + 1 = 9 - 6\sqrt{x+8} + x + 8 - 18 + 6\sqrt{x+8} + 1 = x$. So $(g\circ g^{-1})(4) = 4$.