1. **State the problem:** We have a curve passing through the point $(1,-5)$ with a gradient function (derivative) given by $\frac{dy}{dx} = 4x^3$. We need to find the value of $x$ when $y=10$. 2. **Use the gradient function to find the original function:** Since $\frac{dy}{dx} = 4x^3$, integrate both sides with respect to $x$ to find $y$: $$y = \int 4x^3 \, dx = 4 \cdot \frac{x^4}{4} + C = x^4 + C$$ 3. **Use the point $(1,-5)$ to find $C$:** Substitute $x=1$ and $y=-5$ into $y = x^4 + C$: $$-5 = 1^4 + C \implies -5 = 1 + C \implies C = -6$$ 4. **Write the full equation of the curve:** $$y = x^4 - 6$$ 5. **Find $x$ when $y=10$:** Substitute $y=10$: $$10 = x^4 - 6 \implies x^4 = 16$$ 6. **Solve for $x$:** $$x = \pm \sqrt[4]{16} = \pm 2$$ **Final answer:** The values of $x$ when $y=10$ are $x=2$ and $x=-2$.