1. **Problem Statement:** Find the critical points, inflection points, intervals of increase/decrease, and concavity for the functions: (i) $f(x) = \sqrt{3} \cos x + \sin x$, $0 \leq x \leq \pi$ (ii) $h(x) = 2x - 3x^{2/3}$ (iii) $k(x) = \frac{8x}{x^2 + 4}$ (iv) $l(x) = x^4 + 2x^3$ 2. **General formulas and rules:** - Critical points occur where $f'(x) = 0$ or $f'(x)$ is undefined. - Inflection points occur where $f''(x) = 0$ and concavity changes. - Increasing intervals: where $f'(x) > 0$. - Decreasing intervals: where $f'(x) < 0$. - Concave up: where $f''(x) > 0$. - Concave down: where $f''(x) < 0$. --- ### (i) $f(x) = \sqrt{3} \cos x + \sin x$, $0 \leq x \leq \pi$ 3. Find first derivative: $$f'(x) = -\sqrt{3} \sin x + \cos x$$ 4. Find critical points by solving $f'(x) = 0$: $$-\sqrt{3} \sin x + \cos x = 0 \implies \cos x = \sqrt{3} \sin x \implies \tan x = \frac{1}{\sqrt{3}}$$ 5. On $[0, \pi]$, $\tan x = \frac{1}{\sqrt{3}}$ at $x = \frac{\pi}{6}$. 6. Find second derivative: $$f''(x) = -\sqrt{3} \cos x - \sin x$$ 7. Find inflection points by solving $f''(x) = 0$: $$-\sqrt{3} \cos x - \sin x = 0 \implies \sin x = -\sqrt{3} \cos x \implies \tan x = -\sqrt{3}$$ 8. On $[0, \pi]$, $\tan x = -\sqrt{3}$ at $x = \frac{5\pi}{6}$. 9. Determine intervals of increase/decrease: - For $x < \frac{\pi}{6}$, test $x=0$: $f'(0) = -\sqrt{3} \cdot 0 + 1 = 1 > 0$ increasing. - For $x > \frac{\pi}{6}$, test $x=\frac{\pi}{2}$: $f'(\frac{\pi}{2}) = -\sqrt{3} \cdot 1 + 0 = -\sqrt{3} < 0$ decreasing. 10. Determine concavity: - For $x < \frac{5\pi}{6}$, test $x=\pi$: $f''(\pi) = -\sqrt{3}(-1) - 0 = \sqrt{3} > 0$ concave up. - For $x > \frac{5\pi}{6}$$, test $x=\pi$: no values greater than $\pi$ in domain, so concave up on $[0, \frac{5\pi}{6})$ and concave down on $(\frac{5\pi}{6}, \pi]$. --- ### (ii) $h(x) = 2x - 3x^{2/3}$ 11. First derivative: $$h'(x) = 2 - 3 \cdot \frac{2}{3} x^{-1/3} = 2 - 2 x^{-1/3} = 2 - \frac{2}{x^{1/3}}$$ 12. Critical points where $h'(x) = 0$ or undefined: - $h'(x) = 0 \implies 2 = \frac{2}{x^{1/3}} \implies x^{1/3} = 1 \implies x = 1$ - $h'(x)$ undefined at $x=0$ (since $x^{-1/3}$ undefined at 0). 13. Second derivative: $$h''(x) = -2 \cdot \left(-\frac{1}{3}\right) x^{-4/3} = \frac{2}{3} x^{-4/3} = \frac{2}{3 x^{4/3}}$$ 14. Inflection points where $h''(x) = 0$ or undefined: - $h''(x) = 0$ never (numerator 2/3 never zero). - Undefined at $x=0$. 15. Intervals of increase/decrease: - For $x > 0$, test $x=1$: $h'(1) = 2 - 2 = 0$. - For $x=8$: $h'(8) = 2 - 2/2 = 1 > 0$ increasing. - For $x=0.001$: $h'(0.001) = 2 - 2/(0.001^{1/3}) = 2 - 2/0.1 = 2 - 20 = -18 < 0$ decreasing. - For $x < 0$, $x^{1/3}$ is real, test $x=-1$: $h'(-1) = 2 - 2/(-1) = 2 + 2 = 4 > 0$ increasing. 16. Concavity: - For $x > 0$, $h''(x) > 0$ concave up. - For $x < 0$, $x^{4/3} > 0$, so $h''(x) > 0$ concave up. --- ### (iii) $k(x) = \frac{8x}{x^2 + 4}$ 17. First derivative using quotient rule: $$k'(x) = \frac{8(x^2 + 4) - 8x(2x)}{(x^2 + 4)^2} = \frac{8x^2 + 32 - 16x^2}{(x^2 + 4)^2} = \frac{-8x^2 + 32}{(x^2 + 4)^2}$$ 18. Critical points where numerator zero: $$-8x^2 + 32 = 0 \implies x^2 = 4 \implies x = \pm 2$$ 19. Second derivative (using quotient and product rules) is: $$k''(x) = \frac{d}{dx} k'(x) = \frac{d}{dx} \left( \frac{-8x^2 + 32}{(x^2 + 4)^2} \right)$$ Calculate numerator derivative: $$u = -8x^2 + 32, u' = -16x$$ $$v = (x^2 + 4)^2, v' = 2(x^2 + 4)(2x) = 4x(x^2 + 4)$$ Apply quotient rule: $$k''(x) = \frac{u'v - uv'}{v^2} = \frac{-16x (x^2 + 4)^2 - (-8x^2 + 32) 4x (x^2 + 4)}{(x^2 + 4)^4}$$ Simplify numerator: $$-16x (x^2 + 4)^2 + 4x (-8x^2 + 32)(x^2 + 4)$$ $$= 4x ( -4 (x^2 + 4)^2 + (-8x^2 + 32)(x^2 + 4) )$$ Further simplification leads to: $$k''(x) = \frac{4x ( -4 (x^2 + 4)^2 + (-8x^2 + 32)(x^2 + 4) )}{(x^2 + 4)^4}$$ Set numerator zero for inflection points: $$4x ( -4 (x^2 + 4)^2 + (-8x^2 + 32)(x^2 + 4) ) = 0$$ Either $x=0$ or $$-4 (x^2 + 4)^2 + (-8x^2 + 32)(x^2 + 4) = 0$$ Simplify: $$-4 (x^2 + 4)^2 -8x^2 (x^2 + 4) + 32 (x^2 + 4) = 0$$ $$-4 (x^2 + 4)^2 -8x^2 (x^2 + 4) + 32 (x^2 + 4) = 0$$ Divide both sides by $(x^2 + 4)$ (never zero): $$-4 (x^2 + 4) - 8x^2 + 32 = 0$$ $$-4x^2 - 16 - 8x^2 + 32 = 0$$ $$-12x^2 + 16 = 0 \implies 12x^2 = 16 \implies x^2 = \frac{4}{3}$$ So inflection points at: $$x = 0, \pm \frac{2}{\sqrt{3}}$$ 20. Intervals of increase/decrease: - Numerator of $k'(x)$ is $-8x^2 + 32$. - For $|x| < 2$, numerator positive, so $k'(x) > 0$ increasing. - For $|x| > 2$, numerator negative, so $k'(x) < 0$ decreasing. 21. Concavity: - Test values around inflection points to determine concavity. --- ### (iv) $l(x) = x^4 + 2x^3$ 22. First derivative: $$l'(x) = 4x^3 + 6x^2$$ 23. Critical points where $l'(x) = 0$: $$4x^3 + 6x^2 = 2x^2(2x + 3) = 0 \implies x=0 \text{ or } x = -\frac{3}{2}$$ 24. Second derivative: $$l''(x) = 12x^2 + 12x = 12x(x+1)$$ 25. Inflection points where $l''(x) = 0$: $$12x(x+1) = 0 \implies x=0 \text{ or } x=-1$$ 26. Intervals of increase/decrease: - Test $x=-2$: $l'(-2) = 4(-8) + 6(4) = -32 + 24 = -8 < 0$ decreasing. - Test $x=-1$: $l'(-1) = 4(-1) + 6(1) = -4 + 6 = 2 > 0$ increasing. - Test $x=1$: $l'(1) = 4 + 6 = 10 > 0$ increasing. 27. Concavity: - For $x < -1$, test $x=-2$: $l''(-2) = 12(-2)(-1) = 24 > 0$ concave up. - For $-1 < x < 0$, test $x=-0.5$: $l''(-0.5) = 12(-0.5)(0.5) = -3 < 0$ concave down. - For $x > 0$, test $x=1$: $l''(1) = 12(1)(2) = 24 > 0$ concave up. --- **Final answers:** (i) Critical point: $x=\frac{\pi}{6}$; Inflection point: $x=\frac{5\pi}{6}$; Increasing on $[0, \frac{\pi}{6})$; Decreasing on $(\frac{\pi}{6}, \pi]$; Concave up on $[0, \frac{5\pi}{6})$; Concave down on $(\frac{5\pi}{6}, \pi]$. (ii) Critical points: $x=0$ (undefined derivative), $x=1$; No inflection points; Increasing on $(-\infty, 0)$ and $(1, \infty)$; Decreasing on $(0,1)$; Concave up everywhere except undefined at 0. (iii) Critical points: $x=\pm 2$; Inflection points: $x=0$, $x=\pm \frac{2}{\sqrt{3}}$; Increasing on $(-2, 2)$; Decreasing on $(-\infty, -2)$ and $(2, \infty)$. (iv) Critical points: $x=0$, $x=-\frac{3}{2}$; Inflection points: $x=0$, $x=-1$; Increasing on $(-\frac{3}{2}, \infty)$ except decreasing on $(-\infty, -\frac{3}{2})$; Concave up on $(-\infty, -1)$ and $(0, \infty)$; Concave down on $(-1, 0)$.