미분가능성과 상수

함수 $f \left( x \right)$가 $x = 2$에서 미분가능하므로 함수 $f \left( x \right)$는 $x = 2$에서 연속이다. $\displaystyle \lim\limits _{x \rightarrow 2 -} f \left( x \right) = \lim\limits _{x \rightarrow 2 +} f \left( x \right) = f \left( 2 \right)$에서 $2 - a = 4 + 2 b + a$, $b = - a - 1$ $\displaystyle \lim\limits _{x \rightarrow 2 -} \frac{f \left( x \right) - f \left( 2 \right)}{x - 2}$$\displaystyle = \lim\limits _{x \rightarrow 2 -} \frac{\left( x - a \right) - \left( 2 - a \right)}{x - 2}$$\displaystyle = \lim\limits _{x \rightarrow 2 -} \frac{x - 2}{x - 2} = 1$ $\displaystyle \lim\limits _{x \rightarrow 2 +} \frac{f \left( x \right) - f \left( 2 \right)}{x - 2}$$\displaystyle = \lim\limits _{x \rightarrow 2 +} \frac{\left( x ^{2} + b x + a \right) - \left( 2 - a \right)}{x - 2}$ $\displaystyle = \lim\limits _{x \rightarrow 2 +} \frac{x ^{2} - \left( a + 1 \right) x + 2 \left( a - 1 \right)}{x - 2}$ $\displaystyle = \lim\limits _{x \rightarrow 2 +} \frac{\left( x - 2 \right) \left( x - a + 1 \right)}{x - 2}$ $\displaystyle = \lim\limits _{x \rightarrow 2 +} {} \left( x - a + 1 \right) = 3 - a$ 함수 $f \left( x \right)$가 $x = 2$에서 미분가능하므로 $1 = 3 - a$에서 $a = 2$, $b = - 3$ 따라서 $f \left( 2 \right) = 0$