1. **State the problem:** Find the stationary points of the function $$f(x) = \frac{x^5}{5} - \frac{13x^3}{3} + 36x - 20$$ Stationary points are points where the first derivative $f'(x)$ equals zero. 2. **Find the first derivative:** $$f'(x) = \frac{d}{dx} \left( \frac{x^5}{5} \right) - \frac{d}{dx} \left( \frac{13x^3}{3} \right) + \frac{d}{dx} (36x) - \frac{d}{dx} (20)$$ Calculate each term: $$\frac{d}{dx} \left( \frac{x^5}{5} \right) = x^4$$ $$\frac{d}{dx} \left( \frac{13x^3}{3} \right) = 13x^2$$ $$\frac{d}{dx} (36x) = 36$$ $$\frac{d}{dx} (20) = 0$$ So, $$f'(x) = x^4 - 13x^2 + 36$$ 3. **Set the first derivative equal to zero to find stationary points:** $$x^4 - 13x^2 + 36 = 0$$ 4. **Substitute $y = x^2$ to solve quadratic in $y$:** $$y^2 - 13y + 36 = 0$$ 5. **Factor the quadratic:** $$y^2 - 13y + 36 = (y - 9)(y - 4) = 0$$ So, $$y = 9 \quad \text{or} \quad y = 4$$ 6. **Recall $y = x^2$, solve for $x$:** For $y=9$: $$x^2 = 9 \implies x = \pm 3$$ For $y=4$: $$x^2 = 4 \implies x = \pm 2$$ 7. **List stationary points:** The stationary points occur at $$x = -3, -2, 2, 3$$ 8. **Optional: Find $f(x)$ values at stationary points (for completeness):** Calculate $f(x)$ for each $x$: - For $x=3$: $$f(3) = \frac{3^5}{5} - \frac{13 \times 3^3}{3} + 36 \times 3 - 20 = \frac{243}{5} - \frac{351}{3} + 108 - 20 = 48.6 -117 +108 -20 = 19.6$$ - For $x=-3$: $$f(-3) = \frac{(-3)^5}{5} - \frac{13 (-3)^3}{3} + 36(-3) - 20 = \frac{-243}{5} - \frac{-351}{3} -108 -20 = -48.6 + 117 -108 -20 = -59.6$$ - For $x=2$: $$f(2) = \frac{32}{5} - \frac{13 \times 8}{3} + 72 - 20 = 6.4 - 34.6667 + 72 - 20 = 23.7333$$ - For $x=-2$: $$f(-2) = \frac{-32}{5} - \frac{13 (-8)}{3} - 72 - 20 = -6.4 + 34.6667 -72 - 20 = -63.7333$$ **Final answer:** The stationary points are at $$(-3, -59.6), (-2, -63.7333), (2, 23.7333), (3, 19.6)$$