QuestionSeptember 20, 2025

Show algebraically that f and g are inverse functions. (Example 3) 30 f(x)=(x^2)/(4)+8,xgeqslant 0 g(x)=sqrt (4x-32)

Show algebraically that f and g are inverse functions. (Example 3) 30 f(x)=(x^2)/(4)+8,xgeqslant 0 g(x)=sqrt (4x-32)
Show algebraically that f and g are inverse functions. (Example 3) 30 f(x)=(x^2)/(4)+8,xgeqslant 0 g(x)=sqrt (4x-32)

Solution
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Answer

f(g(x)) = x and g(f(x)) = x, so f and g are inverse functions. Explanation 1. Compose f(g(x)) Substitute g(x) into f(x): f(g(x)) = \frac{(\sqrt{4x-32})^2}{4} + 8. 2. Simplify f(g(x)) (\sqrt{4x-32})^2 = 4x-32, so f(g(x)) = \frac{4x-32}{4} + 8 = x - 8 + 8 = x. 3. Compose g(f(x)) Substitute f(x) into g(x): g(f(x)) = \sqrt{4\left(\frac{x^2}{4}+8\right)-32}. 4. Simplify g(f(x)) 4\left(\frac{x^2}{4}+8\right) = x^2 + 32, so g(f(x)) = \sqrt{x^2 + 32 - 32} = \sqrt{x^2} = x (since x \geq 0).

Explanation

1. Compose $f(g(x))$<br /> Substitute $g(x)$ into $f(x)$: $f(g(x)) = \frac{(\sqrt{4x-32})^2}{4} + 8$.<br />2. Simplify $f(g(x))$<br /> $(\sqrt{4x-32})^2 = 4x-32$, so $f(g(x)) = \frac{4x-32}{4} + 8 = x - 8 + 8 = x$.<br />3. Compose $g(f(x))$<br /> Substitute $f(x)$ into $g(x)$: $g(f(x)) = \sqrt{4\left(\frac{x^2}{4}+8\right)-32}$.<br />4. Simplify $g(f(x))$<br /> $4\left(\frac{x^2}{4}+8\right) = x^2 + 32$, so $g(f(x)) = \sqrt{x^2 + 32 - 32} = \sqrt{x^2} = x$ (since $x \geq 0$).
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