1. **State the problem:** We are given the polynomial function $$f(x) = \frac{15}{4}x^4 + 10x^3 - 60x^2 + 30$$ and need to find its local extrema (local minima and maxima). 2. **Formula and rules:** To find local extrema, we first find the critical points by setting the first derivative $$f'(x)$$ equal to zero. Then, we use the second derivative test to classify each critical point. 3. **Find the first derivative:** $$f'(x) = \frac{d}{dx} \left( \frac{15}{4}x^4 + 10x^3 - 60x^2 + 30 \right) = 15x^3 + 30x^2 - 120x$$ 4. **Set the first derivative equal to zero to find critical points:** $$15x^3 + 30x^2 - 120x = 0$$ 5. **Factor out common terms:** $$15x(x^2 + 2x - 8) = 0$$ 6. **Solve each factor:** - $$15x = 0 \Rightarrow x = 0$$ - Solve quadratic $$x^2 + 2x - 8 = 0$$ using the quadratic formula: $$x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 32}}{2} = \frac{-2 \pm \sqrt{36}}{2} = \frac{-2 \pm 6}{2}$$ 7. **Calculate roots:** - $$x = \frac{-2 + 6}{2} = 2$$ - $$x = \frac{-2 - 6}{2} = -4$$ 8. **Critical points are at $$x = -4, 0, 2$$** 9. **Find the second derivative:** $$f''(x) = \frac{d}{dx} (15x^3 + 30x^2 - 120x) = 45x^2 + 60x - 120$$ 10. **Evaluate the second derivative at each critical point:** - At $$x = -4$$: $$f''(-4) = 45(-4)^2 + 60(-4) - 120 = 45 \cdot 16 - 240 - 120 = 720 - 360 = 360 > 0$$ (local minimum) - At $$x = 0$$: $$f''(0) = 0 + 0 - 120 = -120 < 0$$ (local maximum) - At $$x = 2$$: $$f''(2) = 45 \cdot 4 + 120 - 120 = 180 + 0 = 180 > 0$$ (local minimum) 11. **Find the function values at critical points:** - $$f(-4) = \frac{15}{4}(-4)^4 + 10(-4)^3 - 60(-4)^2 + 30 = \frac{15}{4} \cdot 256 - 640 - 960 + 30 = 960 - 640 - 960 + 30 = -610$$ - $$f(0) = 30$$ - $$f(2) = \frac{15}{4} \cdot 16 + 10 \cdot 8 - 60 \cdot 4 + 30 = 60 + 80 - 240 + 30 = -70$$ **Final answer:** - Local minima at $$(-4, -610)$$ and $$(2, -70)$$ - Local maximum at $$(0, 30)$$