The graph of is shown below. State the natural domain of . Does have an inverse function? Explain…

The function is $f(x) = x - 6\sqrt{x} + 1$. For the square root $\sqrt{x}$ to be defined in real numbers, we must have $x \ge 0$. Therefore, the natural domain of $f(x)$ is $[0, \infty)$.

No, $f(x)$ does not have an inverse function on its natural domain. The graph fails the horizontal line test because a horizontal line intersects the curve at more than one point, so $f(x)$ is not one-to-one. Algebraically, $y = (\sqrt{x})^2 - 6\sqrt{x} + 1 \Rightarrow (\sqrt{x})^2 - 6\sqrt{x} + (1 - y) = 0$. Solving for $\sqrt{x}$: $\sqrt{x} = \frac{6 \pm \sqrt{36 - 4(1-y)}}{2} = 3 \pm \sqrt{y + 8}$, yielding two values of $x$ for each $y$ in the range.

For $g(x) = x - 6\sqrt{x} + 1$ on $[0, 9]$, $\sqrt{x} \le 3$, so we choose the negative branch: $\sqrt{x} = 3 - \sqrt{y + 8} \Rightarrow g^{-1}(x) = (3 - \sqrt{x + 8})^2$. Verification: $(g \circ g^{-1})(4) = g(g^{-1}(4))$. $g^{-1}(4) = (3 - \sqrt{4 + 8})^2 = (3 - \sqrt{12})^2 = 9 - 6\sqrt{12} + 12 = 21 - 12\sqrt{3}$. $g(g^{-1}(4)) = (3 - \sqrt{x+8})^2 - 6(3 - \sqrt{x+8}) + 1 = 9 - 6\sqrt{x+8} + x + 8 - 18 + 6\sqrt{x+8} + 1 = x$. At $x = 4$, $(g \circ g^{-1})(4) = 4$.

For $h(x) = x - 6\sqrt{x} + 1$ on $[9, \infty)$, $\sqrt{x} \ge 3$, so we choose the positive branch: $\sqrt{x} = 3 + \sqrt{y + 8} \Rightarrow h^{-1}(x) = (3 + \sqrt{x + 8})^2$. Verification: $(h \circ h^{-1})(x) = (3 + \sqrt{x + 8})^2 - 6(3 + \sqrt{x + 8}) + 1 = 9 + 6\sqrt{x + 8} + x + 8 - 18 - 6\sqrt{x + 8} + 1 = x$. At $x = 16$, $(h \circ h^{-1})(16) = 16$.

Setting $g^{-1}(x) = h^{-1}(x)$: $(3 - \sqrt{x + 8})^2 = (3 + \sqrt{x + 8})^2$ $9 - 6\sqrt{x + 8} + x + 8 = 9 + 6\sqrt{x + 8} + x + 8$ $-6\sqrt{x + 8} = 6\sqrt{x + 8} \Rightarrow 12\sqrt{x + 8} = 0 \Rightarrow \sqrt{x + 8} = 0 \Rightarrow x + 8 = 0 \Rightarrow x = -8$.