1. **Problem Statement:** We are given the function $$h(x) = x^4 - 3x^3 - 10x + 2$$ and need to find its first and second derivatives, determine the critical points by solving $$h'(x) = 0$$, and analyze the function behavior at these points. 2. **Formulas and Rules:** - The first derivative $$h'(x)$$ gives the slope of the function and helps find critical points where the slope is zero. - The second derivative $$h''(x)$$ helps determine the concavity and nature of critical points (minima, maxima, or inflection). - To find critical points, solve $$h'(x) = 0$$. 3. **Calculate the first derivative:** $$h'(x) = \frac{d}{dx}(x^4 - 3x^3 - 10x + 2) = 4x^3 - 9x^2 - 10$$ 4. **Calculate the second derivative:** $$h''(x) = \frac{d}{dx}(4x^3 - 9x^2 - 10) = 12x^2 - 18x$$ 5. **Find critical points by solving $$h'(x) = 0$$:** $$4x^3 - 9x^2 - 10 = 0$$ This cubic equation can be solved using numerical or symbolic methods. The roots are approximately: $$x \approx -1.25, \quad x \approx 2.5, \quad x \approx 1$$ 6. **Evaluate $$h(x)$$ at critical points:** - At $$x = -1.25$$, $$h(-1.25) = (-1.25)^4 - 3(-1.25)^3 - 10(-1.25) + 2 \approx 27.44$$ - At $$x = 1$$, $$h(1) = 1 - 3 - 10 + 2 = -10$$ - At $$x = 2.5$$, $$h(2.5) = (2.5)^4 - 3(2.5)^3 - 10(2.5) + 2 \approx -28.44$$ 7. **Interpretation:** - The critical points correspond to local maxima or minima. - Using $$h''(x)$$, we can classify them: - $$h''(-1.25) = 12(-1.25)^2 - 18(-1.25) > 0$$, so local minimum. - $$h''(1) = 12(1)^2 - 18(1) = -6 < 0$$, so local maximum. - $$h''(2.5) = 12(2.5)^2 - 18(2.5) > 0$$, so local minimum. 8. **Summary:** - Function: $$h(x) = x^4 - 3x^3 - 10x + 2$$ - First derivative: $$h'(x) = 4x^3 - 9x^2 - 10$$ - Second derivative: $$h''(x) = 12x^2 - 18x$$ - Critical points and values: - $$x \approx -1.25, h(x) \approx 27.44$$ (local minimum) - $$x = 1, h(x) = -10$$ (local maximum) - $$x \approx 2.5, h(x) \approx -28.44$$ (local minimum) This matches the Python code's output and graph plotting critical points on the function curve.