1. Find the derivatives of the following functions: 1.a. Given $F(x) = 2x^2(3x^4 - 2)$, use the product rule: $\frac{d}{dx}[u v] = u' v + u v'$. Here, $u = 2x^2$, $v = 3x^4 - 2$. Calculate derivatives: $u' = 4x$, $v' = 12x^3$. Then, $$F'(x) = 4x(3x^4 - 2) + 2x^2(12x^3) = 12x^5 - 8x + 24x^5 = 36x^5 - 8x$$ 1.b. For $F(x) = (2x - 9)(x^2 + 6)$, again use product rule. Let $u = 2x - 9$, $v = x^2 + 6$. Derivatives: $u' = 2$, $v' = 2x$. Then, $$F'(x) = 2(x^2 + 6) + (2x - 9)(2x) = 2x^2 + 12 + 4x^2 - 18x = 6x^2 - 18x + 12$$ 1.c. For $Y = \frac{t^2 - t}{t^3 - t}$, use quotient rule: $$Y' = \frac{(t^3 - t)(2t - 1) - (t^2 - t)(3t^2 - 1)}{(t^3 - t)^2}$$ Simplify numerator: $(t^3 - t)(2t - 1) = 2t^4 - t^3 - 2t^2 + t$ $(t^2 - t)(3t^2 - 1) = 3t^4 - t^2 - 3t^3 + t$ Numerator: $[2t^4 - t^3 - 2t^2 + t] - [3t^4 - t^2 - 3t^3 + t] = -t^4 + 2t^3 - t^2$ So, $$Y' = \frac{-t^4 + 2t^3 - t^2}{(t^3 - t)^2} = \frac{-t^2(t^2 - 2t + 1)}{(t^3 - t)^2} = \frac{-t^2(t - 1)^2}{(t^3 - t)^2}$$ 1.d. For $Y = \left(\frac{2x + 3}{3x - 1}\right)^3$, use chain and quotient rules. Let $u = \frac{2x + 3}{3x - 1}$, then $Y = u^3$. Derivative: $$Y' = 3u^2 \cdot u'$$ Find $u'$: $$u' = \frac{(3x - 1)(2) - (2x + 3)(3)}{(3x - 1)^2} = \frac{6x - 2 - 6x - 9}{(3x - 1)^2} = \frac{-11}{(3x - 1)^2}$$ Therefore, $$Y' = 3 \left(\frac{2x + 3}{3x - 1}\right)^2 \cdot \frac{-11}{(3x - 1)^2} = -33 \frac{(2x + 3)^2}{(3x - 1)^4}$$ 1.e. For $Y = \sqrt{4 - w} = (4 - w)^{1/2}$, use chain rule: $$Y' = \frac{1}{2}(4 - w)^{-1/2} \cdot (-1) = -\frac{1}{2\sqrt{4 - w}}$$ 2. Investigate successive derivatives: 2.a. $y = x^3 + 3x^2 + 9x + 7$ First derivative: $$y' = 3x^2 + 6x + 9$$ Second derivative: $$y'' = 6x + 6$$ Third derivative: $$y''' = 6$$ Higher derivatives are zero. 2.b. $y = (5 - x)^4$ First derivative: $$y' = 4(5 - x)^3 \cdot (-1) = -4(5 - x)^3$$ Second derivative: $$y'' = -4 \cdot 3(5 - x)^2 \cdot (-1) = 12(5 - x)^2$$ Third derivative: $$y''' = 12 \cdot 2(5 - x) \cdot (-1) = -24(5 - x)$$ Fourth derivative: $$y^{(4)} = -24 \cdot (-1) = 24$$ Higher derivatives are zero. 3. For each function, find critical values, test concavity, find inflection points, and describe graph shape. 3.a. $f(x) = x^3 - 18x^2 + 96x - 80$ First derivative: $$f'(x) = 3x^2 - 36x + 96$$ Set $f'(x) = 0$: $$3x^2 - 36x + 96 = 0 \Rightarrow x^2 - 12x + 32 = 0$$ Solve quadratic: $$x = \frac{12 \pm \sqrt{144 - 128}}{2} = \frac{12 \pm 4}{2}$$ Critical points: $$x = 8, \quad x = 4$$ Second derivative: $$f''(x) = 6x - 36$$ Evaluate at critical points: $$f''(8) = 48 - 36 = 12 > 0 \Rightarrow \text{local minimum at } x=8$$ $$f''(4) = 24 - 36 = -12 < 0 \Rightarrow \text{local maximum at } x=4$$ Find inflection points by setting $f''(x) = 0$: $$6x - 36 = 0 \Rightarrow x = 6$$ 3.b. $y = -9x^2 + 72x - 13$ First derivative: $$y' = -18x + 72$$ Set $y' = 0$: $$-18x + 72 = 0 \Rightarrow x = 4$$ Second derivative: $$y'' = -18$$ Since $y'' < 0$ always, the function is concave down everywhere. At $x=4$, $y$ has a local maximum. No inflection points since $y''$ is constant. Slug: "derivatives critical" Subject: "calculus" Desmos: {"latex": "f(x) = x^3 - 18x^2 + 96x - 80", "features": {"intercepts": true, "extrema": true}} q_count: 7