1. The problem asks to find the exact values of $x$ when the gradient of the curve $y = \tan^{-1}(4x)$ is $\frac{1}{4}$. 2. The formula for the gradient of $y = \tan^{-1}(u)$ is $\frac{dy}{dx} = \frac{1}{1+u^2} \cdot \frac{du}{dx}$. Here, $u = 4x$, so $\frac{du}{dx} = 4$. 3. Substitute into the formula: $$\frac{dy}{dx} = \frac{1}{1+(4x)^2} \times 4 = \frac{4}{1+16x^2}.$$ 4. Set the gradient equal to $\frac{1}{4}$: $$\frac{4}{1+16x^2} = \frac{1}{4}.$$ 5. Cross-multiply and solve for $x$: $$16 = 1 + 16x^2 \implies 16x^2 = 15 \implies x^2 = \frac{15}{16} \implies x = \pm \frac{\sqrt{15}}{4}.$$ **Final answer:** $$x = \pm \frac{\sqrt{15}}{4}.$$