QuestionMarch 21, 2026

Are g(x) and f(x) Inverse functions on the set of x-values where their compositions are defined? f(x)=(-4x+2)/(-3x-3) g(x)=(3x+2)/(-3x+4)

Are g(x) and f(x) Inverse functions on the set of x-values where their compositions are defined? f(x)=(-4x+2)/(-3x-3) g(x)=(3x+2)/(-3x+4)
Are g(x) and f(x) Inverse functions on the set of x-values where their compositions are
defined?
f(x)=(-4x+2)/(-3x-3)
g(x)=(3x+2)/(-3x+4)

Solution
4.3(179 votes)

Answer

Yes, f(x) and g(x) are inverse functions on their common domain. Explanation 1. Compute f(g(x)) Substitute g(x) into f(x): f(g(x)) = \frac{-4\left(\frac{3x+2}{-3x+4}\right)+2}{-3\left(\frac{3x+2}{-3x+4}\right)-3} Simplify numerator: \frac{-12x-8}{-3x+4} + 2 = \frac{-12x-8+2(-3x+4)}{-3x+4} = \frac{-12x-8-6x+8}{-3x+4} = \frac{-18x}{-3x+4} Simplify denominator: \frac{-9x-6}{-3x+4} - 3 = \frac{-9x-6+(-3)(-3x+4)}{-3x+4} = \frac{-9x-6+9x-12}{-3x+4} = \frac{-18}{-3x+4} f(g(x)) = \frac{\frac{-18x}{-3x+4}}{\frac{-18}{-3x+4}} = x 2. Compute g(f(x)) Substitute f(x) into g(x): g(f(x)) = \frac{3\left(\frac{-4x+2}{-3x-3}\right)+2}{-3\left(\frac{-4x+2}{-3x-3}\right)+4} Numerator: \frac{-12x+6}{-3x-3} + 2 = \frac{-12x+6+2(-3x-3)}{-3x-3} = \frac{-12x+6-6x-6}{-3x-3} = \frac{-18x}{-3x-3} Denominator: \frac{12x-6}{-3x-3} + 4 = \frac{12x-6+4(-3x-3)}{-3x-3} = \frac{12x-6-12x-12}{-3x-3} = \frac{-18}{-3x-3} g(f(x)) = \frac{\frac{-18x}{-3x-3}}{\frac{-18}{-3x-3}} = x 3. Conclusion Both compositions f(g(x)) and g(f(x)) equal x, so f and g are inverses wherever defined.

Explanation

1. Compute $f(g(x))$<br /> Substitute $g(x)$ into $f(x)$: <br />$f(g(x)) = \frac{-4\left(\frac{3x+2}{-3x+4}\right)+2}{-3\left(\frac{3x+2}{-3x+4}\right)-3}$ <br />Simplify numerator: $ \frac{-12x-8}{-3x+4} + 2 = \frac{-12x-8+2(-3x+4)}{-3x+4} = \frac{-12x-8-6x+8}{-3x+4} = \frac{-18x}{-3x+4}$ <br />Simplify denominator: $\frac{-9x-6}{-3x+4} - 3 = \frac{-9x-6+(-3)(-3x+4)}{-3x+4} = \frac{-9x-6+9x-12}{-3x+4} = \frac{-18}{-3x+4}$ <br />$f(g(x)) = \frac{\frac{-18x}{-3x+4}}{\frac{-18}{-3x+4}} = x$<br /><br />2. Compute $g(f(x))$<br /> Substitute $f(x)$ into $g(x)$: <br />$g(f(x)) = \frac{3\left(\frac{-4x+2}{-3x-3}\right)+2}{-3\left(\frac{-4x+2}{-3x-3}\right)+4}$ <br />Numerator: $\frac{-12x+6}{-3x-3} + 2 = \frac{-12x+6+2(-3x-3)}{-3x-3} = \frac{-12x+6-6x-6}{-3x-3} = \frac{-18x}{-3x-3}$ <br />Denominator: $\frac{12x-6}{-3x-3} + 4 = \frac{12x-6+4(-3x-3)}{-3x-3} = \frac{12x-6-12x-12}{-3x-3} = \frac{-18}{-3x-3}$ <br />$g(f(x)) = \frac{\frac{-18x}{-3x-3}}{\frac{-18}{-3x-3}} = x$<br /><br />3. Conclusion<br /> Both compositions $f(g(x))$ and $g(f(x))$ equal $x$, so $f$ and $g$ are inverses wherever defined.
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