1. Problem: Given $g(x) = \frac{x + 2}{5x - 1}$ with $x \neq \frac{1}{5}$, find the inverse function $g^{-1}(x)$ and determine if $g$ is one-one and onto. 2. To find the inverse, start by setting $y = g(x)$: $$y = \frac{x + 2}{5x - 1}$$ 3. Swap $x$ and $y$ to find the inverse function: $$x = \frac{y + 2}{5y - 1}$$ 4. Solve for $y$: Multiply both sides by $(5y - 1)$: $$x(5y - 1) = y + 2$$ $$5xy - x = y + 2$$ 5. Rearrange terms to isolate $y$: $$5xy - y = x + 2$$ $$y(5x - 1) = x + 2$$ 6. Divide both sides by $(5x - 1)$: $$y = \frac{x + 2}{5x - 1}$$ 7. Notice this is the same as the original function, so the inverse is: $$g^{-1}(x) = \frac{x + 2}{5x - 1}$$ 8. Check if $g$ is one-one (injective): Since $g$ is a rational function with a linear numerator and denominator and the denominator is not zero at $x=\frac{1}{5}$, it is strictly monotonic on intervals separated by $x=\frac{1}{5}$, so $g$ is one-one on its domain. 9. Check if $g$ is onto (surjective): The function can take all real values except possibly the value that makes the denominator zero in the inverse, which is $x=\frac{1}{5}$. Hence, $g$ is onto its codomain excluding that point. Final answer: $$g^{-1}(x) = \frac{x + 2}{5x - 1}$$ "slug": "inverse g", "subject": "algebra", "desmos": {"latex": "y=\frac{x+2}{5x-1}","features": {"intercepts": true,"extrema": false}}, "q_count": 4