1. **State the problem:** Find the point on the curve $y = -2x^4$ where the tangent line is perpendicular to the line $x - y + 1 = 0$. 2. **Rewrite the given line in slope-intercept form:** The line is $x - y + 1 = 0$. Rearranged: $$y = x + 1$$ So, the slope of this line is $m_1 = 1$. 3. **Find the slope of the tangent line to the curve:** The curve is $y = -2x^4$. Differentiate with respect to $x$: $$\frac{dy}{dx} = -2 \cdot 4x^3 = -8x^3$$ So, the slope of the tangent line at any point $x$ is $m_2 = -8x^3$. 4. **Condition for perpendicular lines:** Two lines are perpendicular if their slopes satisfy: $$m_1 \cdot m_2 = -1$$ Substitute $m_1 = 1$ and $m_2 = -8x^3$: $$1 \cdot (-8x^3) = -1$$ Simplify: $$-8x^3 = -1$$ 5. **Solve for $x$:** $$-8x^3 = -1$$ $$\cancel{-8}x^3 = \cancel{-1} \quad \Rightarrow \quad x^3 = \frac{1}{8}$$ Take cube root: $$x = \sqrt[3]{\frac{1}{8}} = \frac{1}{2}$$ 6. **Find the corresponding $y$ coordinate:** Substitute $x = \frac{1}{2}$ into the curve equation: $$y = -2 \left( \frac{1}{2} \right)^4 = -2 \cdot \frac{1}{16} = -\frac{1}{8}$$ 7. **Final answer:** The point on the curve where the tangent line is perpendicular to the line $x - y + 1 = 0$ is: $$\boxed{\left( \frac{1}{2}, -\frac{1}{8} \right)}$$