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Graph of f(x)
Answer the following.
Let . Find and verify .
Let $h(x) = x - 6\sqrt{x} + 1, x \ge 9$. Find $h^{-1}(x)$ and verify $(h \circ h^{-1})(16) = 16$.
Answer
Answer
AIFor $h(x) = x - 6\sqrt{x} + 1$ 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 $x = 16$, $(h \circ h^{-1})(16) = 16$.
From ISC 2027 Specimen Mathematics Paper 1, question 16(iv).