Let h(x) = x - 6√x + 1, x ≥ 9. Find h⁻¹(x) and verify (h ∘ h⁻¹)(16) = 16.

Let $h(x) = x - 6\sqrt{x} + 1, x \ge 9$. Find $h^{-1}(x)$ and verify $(h \circ h^{-1})(16) = 16$.

Answer

Official answer key

On $x \ge 9$, $\sqrt{x} \ge 3$, so $\sqrt{x} = 3 + \sqrt{y+8}$ and $h^{-1}(x) = \left(3 + \sqrt{x+8}\right)^2$. Verification: $h\left(h^{-1}(x)\right) = \left(3 + \sqrt{x+8}\right)^2 - 6\left(3 + \sqrt{x+8}\right) + 1 = 9 + 6\sqrt{x+8} + x + 8 - 18 - 6\sqrt{x+8} + 1 = x$. So $(h\circ h^{-1})(16) = 16$.