1. **State the problem:** We are given the derivative of a function $f'(x) = e^{-x^2} (x^2 - 1)(2x - 3)$ and asked to find all values of $x$ where the tangent line to $f(x)$ is horizontal. 2. **Recall the rule:** The tangent line to $f(x)$ is horizontal where the derivative $f'(x) = 0$. 3. **Set the derivative equal to zero:** $$e^{-x^2} (x^2 - 1)(2x - 3) = 0$$ 4. **Analyze each factor:** - $e^{-x^2}$ is never zero for any real $x$ because the exponential function is always positive. - So, the zeros come from $(x^2 - 1)(2x - 3) = 0$. 5. **Solve each factor:** - $x^2 - 1 = 0 \implies x^2 = 1 \implies x = \pm 1$ - $2x - 3 = 0 \implies 2x = 3 \implies x = \frac{3}{2}$ 6. **List all values:** The tangent line is horizontal at $x = -1$, $x = 1$, and $x = \frac{3}{2}$. **Final answer:** $$x = -1, \quad x = 1, \quad x = \frac{3}{2}$$