1. **Problem 7:** Find the minimum value of the function $g(x) = x^5 - 5x^3 - 20x$ on the interval $[0,3]$. 2. To find the minimum, we first find the critical points by computing the derivative and setting it equal to zero. 3. The derivative is: $$g'(x) = 5x^4 - 15x^2 - 20$$ 4. Set the derivative equal to zero to find critical points: $$5x^4 - 15x^2 - 20 = 0$$ 5. Divide both sides by 5: $$\cancel{5}x^4 - 3x^2 - 4 = 0$$ 6. Let $u = x^2$, then the equation becomes: $$u^2 - 3u - 4 = 0$$ 7. Factor the quadratic: $$(u - 4)(u + 1) = 0$$ 8. So, $u = 4$ or $u = -1$. Since $u = x^2$ and $x^2$ cannot be negative, discard $u = -1$. 9. Thus, $x^2 = 4$ which gives $x = \pm 2$. On the interval $[0,3]$, only $x=2$ is relevant. 10. Evaluate $g(x)$ at critical point and endpoints: - $g(0) = 0^5 - 5\cdot0^3 - 20\cdot0 = 0$ - $g(2) = 2^5 - 5\cdot2^3 - 20\cdot2 = 32 - 40 - 40 = -48$ - $g(3) = 3^5 - 5\cdot3^3 - 20\cdot3 = 243 - 135 - 60 = 48$ 11. The minimum value on $[0,3]$ is $-48$ at $x=2$. --- 12. **Problem 9:** Given $h(x) = \tan^{-1}\left(\frac{x}{2}\right)$, find $h'(x)$. 13. Recall the derivative formula for inverse tangent: $$\frac{d}{dx} \tan^{-1}(u) = \frac{u'}{1 + u^2}$$ 14. Here, $u = \frac{x}{2}$, so: $$u' = \frac{d}{dx} \left(\frac{x}{2}\right) = \frac{1}{2}$$ 15. Substitute into the formula: $$h'(x) = \frac{\frac{1}{2}}{1 + \left(\frac{x}{2}\right)^2} = \frac{\frac{1}{2}}{1 + \frac{x^2}{4}} = \frac{\frac{1}{2}}{\frac{4 + x^2}{4}}$$ 16. Simplify the complex fraction: $$h'(x) = \frac{1}{2} \times \frac{4}{4 + x^2} = \frac{2}{4 + x^2}$$ 17. Therefore, the derivative is: $$h'(x) = \frac{2}{4 + x^2}$$ **Final answers:** - Problem 7 minimum value: $-48$ - Problem 9 derivative: $\frac{2}{4 + x^2}$