1. **Problem Statement:** Find a root of the function $f(x) = x^3 - x - 3$ in the given intervals. 2. **Recall the Intermediate Value Theorem:** If $f$ is continuous on $[a,b]$ and $f(a)$ and $f(b)$ have opposite signs, then there exists at least one root in $(a,b)$. 3. **Evaluate $f(x)$ at the interval endpoints:** - For $(1,4)$: $$f(1) = 1^3 - 1 - 3 = 1 - 1 - 3 = -3$$ $$f(4) = 4^3 - 4 - 3 = 64 - 4 - 3 = 57$$ Since $f(1) < 0$ and $f(4) > 0$, there is a root in $(1,4)$. - For $(-2,-1)$: $$f(-2) = (-2)^3 - (-2) - 3 = -8 + 2 - 3 = -9$$ $$f(-1) = (-1)^3 - (-1) - 3 = -1 + 1 - 3 = -3$$ Both negative, no root guaranteed. - For $(0,1)$: $$f(0) = 0 - 0 - 3 = -3$$ $$f(1) = -3$$ Both negative, no root guaranteed. - For $(-1,0)$: $$f(-1) = -3$$ $$f(0) = -3$$ Both negative, no root guaranteed. 4. **Conclusion:** The only interval where $f(x)$ changes sign is $(1,4)$, so the root lies there. **Final answer:** $f(x)$ has a root in the interval $(1,4)$.