1. **State the problem:** Given functions $f(x) = x^2 - 2x$ and $g(x) = 2x + 3$, and the equation $f \circ g^{-1}(x) = 3$, find the value of $x$. 2. **Find the inverse of $g$: ** Given $y = g(x) = 2x + 3$, solve for $x$: $$y = 2x + 3 \implies 2x = y - 3 \implies x = \frac{y-3}{2}.$$ Thus, $$g^{-1}(x) = \frac{x-3}{2}.$$ 3. **Compute $f(g^{-1}(x))$: ** $$f(g^{-1}(x)) = f\left(\frac{x-3}{2}\right) = \left(\frac{x-3}{2}\right)^2 - 2\left(\frac{x-3}{2}\right).$$ Simplify step by step: $$\left(\frac{x-3}{2}\right)^2 = \frac{(x-3)^2}{4} = \frac{x^2 - 6x + 9}{4}$$ $$- 2 \times \frac{x-3}{2} = -(x - 3) = -x + 3.$$ So, $$f(g^{-1}(x)) = \frac{x^2 - 6x + 9}{4} - x + 3.$$ 4. **Set equation equal to 3 and solve for $x$: ** $$\frac{x^2 - 6x + 9}{4} - x + 3 = 3.$$ Subtract 3 from both sides: $$\frac{x^2 - 6x + 9}{4} - x = 0.$$ Multiply both sides by 4: $$x^2 - 6x + 9 - 4x = 0 \implies x^2 - 10x + 9 = 0.$$ 5. **Solve quadratic equation $x^2 - 10x + 9 = 0$: ** Use quadratic formula: $$x = \frac{10 \pm \sqrt{(-10)^2 - 4 \times 1 \times 9}}{2} = \frac{10 \pm \sqrt{100 - 36}}{2} = \frac{10 \pm \sqrt{64}}{2} = \frac{10 \pm 8}{2}.$$ Two solutions: $$x = \frac{10 + 8}{2} = 9,$$ $$x = \frac{10 - 8}{2} = 1.$$ **Final answer:** $x = 1$ or $x = 9$.