미분가능 조건

함수 $f ( x )$가 $x = 1$에서 미분가능하므로 $x = 1$에서 연속이다. $\displaystyle \lim\limits _{x \rightarrow 1 -} {} f \left( x \right) = \lim\limits _{x \rightarrow 1 +} {} f \left( x \right) = f \left( 1 \right)$ $\displaystyle \lim\limits _{x \rightarrow 1 -} {} f \left( x \right) = \lim\limits _{x \rightarrow 1 -} {} \left( 3 x + a \right) = a + 3$ $\displaystyle \lim\limits _{x \rightarrow 1 +} {} f \left( x \right) = \lim\limits _{x \rightarrow 1 +} {} \left( 2 x ^{3} + bx + 1 \right) = b + 3$ $f \left( 1 \right) = a + 3$ $a + 3 = b + 3$, $a = b$ 함수 $f ( x )$가 $x = 1$에서 미분가능하므로 $\displaystyle \lim\limits _{x \rightarrow 1 -} {} \frac{f \left( x \right) - f \left( 1 \right)}{x - 1} = \lim\limits _{x \rightarrow 1 +} {} \frac{f \left( x \right) - f \left( 1 \right)}{x - 1}$ $\displaystyle \lim\limits _{x \rightarrow 1 -} {} \frac{f \left( x \right) - f \left( 1 \right)}{x - 1} = \lim\limits _{x \rightarrow 1 -} {} \frac{3 x + a - \left( a + 3 \right)}{x - 1} = 3$ $\displaystyle \lim\limits _{x \rightarrow 1 +} \frac{f ( x ) - f ( 1 )}{x - 1}$$\displaystyle = \lim\limits _{x \rightarrow 1 +} \frac{\left( 2 x ^{3} + a x + 1 \right) - ( a + 3 )}{x - 1}$ $\displaystyle = \lim\limits _{x \rightarrow 1 +} \frac{( x - 1 ) \left( 2 x ^{2} + 2 x + a + 2 \right)}{x - 1}$ $= a + 6$ $3 = a + 6$, $a = - 3$, $b = - 3$ 따라서 $a + b = - 6$