삼차함수와 이차함수

$f ( \alpha ) = g ( \alpha )$이고 $f' ( \alpha ) = g' ( \alpha )$이므로 $f ( x ) - g ( x ) = ( x - \alpha ) ^{2} ( x - k )$라 하면 $\displaystyle f' ( x ) - g' ( x ) = 3 ( x - \alpha ) \left( x - \frac{2 k + \alpha}{3} \right)$ … ㉠ 한편 $f' ( \alpha ) = g' ( \alpha )$, $f' ( \beta ) = g' ( \beta )$이므로 $f' ( x ) - g' ( x ) = 3 ( x - \alpha ) ( x - \beta )$ … ㉡ ㉠, ㉡에서 $\displaystyle \frac{2 k + \alpha}{3} = \beta$ $\displaystyle k = \frac{3 \beta - \alpha}{2}$ $\displaystyle \therefore f ( x ) - g ( x ) = ( x - \alpha ) ^{2} \left( x - \frac{3 \beta - \alpha}{2} \right)$ … ㉢ $g' ( \alpha ) = - 16$, $g' ( \beta ) = 16$이므로 $\displaystyle g ( x ) = 2 \left( x - \frac{\alpha + \beta}{2} \right) ^{2} + c$ $\displaystyle g' ( x ) = 4 \left( x - \frac{\alpha + \beta}{2} \right)$ $g' ( \beta ) = 16$에서 $\beta - \alpha = 8$ … ㉣ 구하는 값은 ㉢, ㉣에 의해 $\displaystyle g ( \beta + 1 ) - f ( \beta + 1 ) = - ( \beta + 1 - \alpha ) ^{2} \left( \beta + 1 - \frac{3 \beta - \alpha}{2} \right)$ $= 243$