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Graph of f(x)
Answer the following.
Let . Find and verify .
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
Answer
AIFor $g(x) = x - 6\sqrt{x} + 1$ 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 $x = 4$, $(g \circ g^{-1})(4) = 4$.
From ISC 2027 Specimen Mathematics Paper 1, question 16(iii).