กำหนดฟังก์ชัน \( \mathrm{f}(\mathrm{x})=2 \mathrm{x}^{3}-9 \mathrm{x}^{2}+12 \mathrm{x}-2 \) จุดต่ำสุดสัมพัทธ์ เท่ากับคือข้อใด \( \begin{array}{ll}\text { 1) }\left(\frac{3}{2}, \frac{5}{2}\right) & \text { 2) }\left(\frac{1}{2}, \frac{3}{2}\right) \\ \text { 3) }(1,3) & \text { 4) }(2,2)\end{array} \)

Shown that \( \int_{1}^{\infty} \frac{1}{x} d x \quad \) is divergent

The sum of the critical numbers of \( f(x)=x \sqrt[3]{4-x} \) is equal to

lopo aver scritto l'equazione della parabola con asse parallelo all'asse \( y \) che passa per i punti dice inate \( (0,1) \), \( (1,2) \mathrm{e}\left(-\frac{1}{2},-\frac{1}{4}\right) \), trova le rette ad essa tangenti uscenti dal punto \( A(1,3) \). Detern oi l'area della regione di piano delimitata dalle tangenti e dalla parabola. \[ y y=-x^{2}+2 x+1 ; y=2 x+1 ; y=-2 x+5 \text {; area } \]

The value of \( c \) that satisfies the conclusion of Rolle's Theorem for \( f(x)=x^{3}-x^{2}-2 x+2 \) over \( [1, \sqrt{2}] \) is equal to

\( \left.\begin{array}{l}\text { it la fonction numérique définie par : } \\ \qquad\left\{\begin{array}{l}f(x)=\sqrt{x^{2}-x}+x+2 \text { si } x<0 \\ f(x)=x+1-\frac{1}{x \sqrt{x}-1}\end{array} \text { si } x \geq 0\right.\end{array}\right\} \)