극한 조건과 접선 절편

$f ( 1 ) + f ( 1 ) + 12 = 0$ ∴ $f ( 1 ) = - 6$ $\displaystyle \lim\limits _{x \rightarrow 1} \frac{f ( x ^{2} ) + f ( x ) + 12}{x - 1}$ $\displaystyle = \lim\limits _{x \rightarrow 1} \left\{ \frac{f ( x ^{2} ) + 6}{x - 1} + \frac{f ( x ) + 6}{x - 1} \right\}$ $\displaystyle = \lim\limits _{x \rightarrow 1} \left\{ \frac{\{ f ( x ^{2} ) - f ( 1 ) \} ( x + 1 )}{( x - 1 ) ( x + 1 )} + \frac{f ( x ) - f ( 1 )}{x - 1} \right\}$ $\displaystyle = \lim\limits _{x \rightarrow 1} \left\{ \frac{\{ f ( x ^{2} ) - f ( 1 ) \} ( x + 1 )}{x ^{2} - 1} + \frac{f ( x ) - f ( 1 )}{x - 1} \right\}$ $= 2 f' ( 1 ) + f' ( 1 )$ $= 3 f' ( 1 ) = 12$ ∴ $f' ( 1 ) = 4$ 따라서 접선의 방정식이 $y = 4 ( x - 1 ) - 6 = 4 x - 10$이므로 $y$절편은 $- 10$이다.