ข้อสอบคณิต 9 สามัญ - ปี 2559

$$\begin{gathered} \mathrm{จากโจทย์} 𝑓\left(\mathrm{x}\right)=\left\{\begin{array}{c}{\left(\mathrm{x}+1\right)}^2-5 \mathrm{เมื่อ} \mathrm{x}<-1\\ -5 \mathrm{เมื่อ} -1\le \mathrm{x}\le 1\\ {\left(\mathrm{x}-1\right)}^2-5 \mathrm{เมื่อ} \mathrm{x}>1\end{array}\right. \\ \mathrm{จะได้} 𝑓\text{'}\left(\mathrm{x}\right)=\left\{\begin{array}{c} 2\left(\mathrm{x}+1\right)\left(1\right) \mathrm{เมื่อ} \mathrm{x}<-1\\ 0 \mathrm{เมื่อ} -1\le \mathrm{x}\le 1\\ 2\left(\mathrm{x}-1\right)\left(1\right) \mathrm{เมื่อ} \mathrm{x}>1\end{array}\right. \end{gathered}$$

$$\begin{gathered} \mathrm{ดังนั้น} \left(𝑓\circ 𝑓\right)\text{'}\left(2\right)=𝑓\text{'}\left(𝑓\left(2\right)\right)·𝑓\text{'}\left(2\right) \\ =𝑓\text{'}\left({\left(2-1\right)}^2-5\right)·2\left(2-1\right) \\ =𝑓\text{'}\left(-4\right)·2 \\ =2\left(-4+1\right)\left(2\right) \\ =2\left(-3\right)\left(2\right) \\ =-12 \end{gathered}$$