ข้อสอบคณิต 9 สามัญ - ปี 2562

$$\begin{gathered} \mathrm{นิยาม} f\left(x\right)=y \to x=f^{-1}\left(y\right) \\ \mathrm{จากโจทย์} f\left(x\right) = x^3+2x+3 \\ x = f^{-1}\left(x^3+2x+3\right) \\ x = g\left(x^3+2x+3\right) \end{gathered}$$

$$\begin{gathered} - diff \mathrm{เทียบ} x \mathrm{ทั้ง} 2 \mathrm{ข้าง} \\ \frac{d}{dx}x = \frac{d}{dx}\left[g\left(x^3+2x+3\right)\right] \\ 1 = g\text{'}(x^3+2x+3)\frac{d}{dx}(x^3+2x+3) \\ 1 = g\text{'}(x^3+2x+3)·(3x^2+2) \\ g\text{'}(x^3+2x+3) = \frac{1}{3x^2+2} \to \boxed{1} \end{gathered}$$

$$\begin{gathered} - \mathrm{ต้องการหาค่า} g\text{'}\left(6\right) \\ \mathrm{แสดงว่า} x^3+2x+3 = 6 \\ x^3+2x-3 = 0 \to \boxed{2} \\ - \mathrm{ลองแทนค่า} x=1 \mathrm{จะได้} 1^3+2\left(1\right)-3 = 0 \\ 0 = 0 \end{gathered}$$

$\begin{array}{cccc}1& 0& 2& -3\\ & 1& 1& 3\end{array}\begin{array}{cccc}1& 1& 3& 0\end{array}1$

$$\begin{gathered} \mathrm{จาก} \boxed{2} \mathrm{จะได้} \left(x-1\right)\left(x^2+x+3\right) = 0 \\ - \mathrm{แต่} x^2+x+3>0 \mathrm{เสมอ} \mathrm{ตัดทิ้งได้} \\ x-1 = 0 \\ x = 1 ; \mathrm{แทนใน} \boxed{1} \end{gathered}$$

$$\begin{gathered} \mathrm{จาก} \boxed{1} g\text{'}\left(x^3+2x+3\right) = \frac{1}{3x^2+2} \\ g\text{'}\left(6\right) = \frac{1}{3{\left(1\right)}^2+2} \\ g\text{'}\left(6\right) = \frac{1}{5} \end{gathered}$$