\(f(x)=x^4+x^3-3x^2+1\)의 변곡점에서 접선의 방정식을 구하라.
풀이
\(\displaystyle f'(x)=4x^3+3x^2-6x=x(4x^2+3x-6)\)
\(\displaystyle f'{}'(x)=12x^2+6x-6=6(2x^2+x-1)=6(2x-1)(x+1)\)
\begin{array} {c|c|c|c|c|c} x & \cdots & -1 & \cdots & \frac{1}{2} & \cdots \\ \hline f'{}’ & + & 0 & – & 0 & + \end{array}
\(\displaystyle f(-1)=(-1)^4+(-1)^3-3(-1)^2+1=1-1-3(1)+1=-2\)
\(\displaystyle f\left(\frac{1}{2}\right)=\left(\frac{1}{2}\right)^4+\left(\frac{1}{2}\right)^3-3\left(\frac{1}{2}\right)^2+1\)
\(\displaystyle=\frac{1}{16}+\frac{1}{8}-\frac{3}{4}+1=\frac{1+2-12+16}{16}=\frac{7}{16}\)
변곡점은 \(\displaystyle\left(-1, -2\right), \left(\frac{1}{2}, \frac{7}{16}\right)\)이다.
\(\displaystyle f'(-1)=(-1)(4(-1)^2+3(-1)-6)\)
\(\displaystyle=-(4(1)+-3-6)=-(4-3-6)=-(-5)=5\)
\(\displaystyle f’\left(\frac{1}{2}\right)=\frac{1}{2}\left(4\left(\frac{1}{2}\right)^2+3\left(\frac{1}{2}\right)-6\right)\)
\(\displaystyle=\frac{1}{2}\left(4\left(\frac{1}{4}\right)+\frac{3}{2}-6\right)=\frac{1}{2}\left(1+\frac{3}{2}-6\right)\)
\(\displaystyle=\frac{1}{2}\left(\frac{2+3-12}{2}\right)=\frac{1}{2}\left(\frac{-7}{2}\right)=-\frac{7}{4}\)
1. 변곡점 \(\displaystyle\left(-1, -2\right)\)에서 접선의 방정식을 구한다.
\(y-f(-1)=f'(-1)(x-(-1))\)
\(y-(-2)=5(x+1)\)
\(y+2=5x+5\)
\(y=5x+3\)
2. 변곡점 \(\displaystyle\left(\frac{1}{2}, \frac{7}{16}\right)\)에서 접선의 방정식을 구한다.
\(\displaystyle y-f\left(\frac{1}{2}\right)=f’\left(\frac{1}{2}\right)\left(x-\frac{1}{2}\right)\)
\(\displaystyle y-\frac{7}{16}=-\frac{7}{4}\left(x-\frac{1}{2}\right)\)
\(\displaystyle y-\frac{7}{16}=-\frac{7}{4}x+\frac{7}{8}\)
\(\displaystyle y=-\frac{7}{4}x+\frac{7}{8}+\frac{7}{16}\)
\(\displaystyle y=-\frac{7}{4}x+\frac{14+7}{16}\)
\(\displaystyle y=-\frac{7}{4}x+\frac{21}{16}\)