1. **State the problem:** We are given the function $f(x) = x^3 - 15x$ and need to determine whether it has maximum or minimum values and understand its impact on network efficiency and scalability. 2. **Find the first derivative:** To find critical points where maxima or minima may occur, compute $$f'(x) = \frac{d}{dx}(x^3 - 15x) = 3x^2 - 15.$$ 3. **Set the first derivative to zero to find critical points:** $$3x^2 - 15 = 0$$ Divide both sides by 3: $$\cancel{3}x^2 - \cancel{3}5 = 0 \Rightarrow x^2 - 5 = 0$$ 4. **Solve for $x$:** $$x^2 = 5 \Rightarrow x = \pm \sqrt{5}.$$ 5. **Find the second derivative to classify critical points:** $$f''(x) = \frac{d}{dx}(3x^2 - 15) = 6x.$$ 6. **Evaluate the second derivative at critical points:** - At $x = \sqrt{5}$: $$f''(\sqrt{5}) = 6\sqrt{5} > 0,$$ so $f$ has a local minimum here. - At $x = -\sqrt{5}$: $$f''(-\sqrt{5}) = 6(-\sqrt{5}) = -6\sqrt{5} < 0,$$ so $f$ has a local maximum here. 7. **Interpretation:** - The function has a local maximum at $x = -\sqrt{5}$ and a local minimum at $x = \sqrt{5}$. - In the context of network traffic, these points could represent thresholds where efficiency or scalability changes. - The maximum point might indicate a peak in network load beyond which performance degrades. - The minimum point might represent a stable, efficient operating point. **Final answer:** The function $f(x) = x^3 - 15x$ has a local maximum at $x = -\sqrt{5}$ and a local minimum at $x = \sqrt{5}$, indicating critical points that impact network efficiency and scalability.