1. The problem is to find $f^{-1}(3)$ given the function $f(x) = x^2 + 2x + 1$. 2. First, understand that $f^{-1}(3)$ means we want to find the value of $x$ such that $f(x) = 3$. 3. Set the equation $x^2 + 2x + 1 = 3$. 4. Simplify the equation: $x^2 + 2x + 1 - 3 = 0$ which gives $x^2 + 2x - 2 = 0$. 5. Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a=1$, $b=2$, and $c=-2$. 6. Calculate the discriminant: $\Delta = 2^2 - 4 \times 1 \times (-2) = 4 + 8 = 12$. 7. Find the roots: $x = \frac{-2 \pm \sqrt{12}}{2} = \frac{-2 \pm 2\sqrt{3}}{2} = -1 \pm \sqrt{3}$. 8. Therefore, $f^{-1}(3)$ has two values: $-1 + \sqrt{3}$ and $-1 - \sqrt{3}$. 9. Note: Since $f(x)$ is a quadratic function, it is not one-to-one over all real numbers, so $f^{-1}(3)$ is not a single value but two values.