곱의 미분계수

$\displaystyle \lim\limits _{x \rightarrow 1} {\frac{f \left( x \right) - a + 2}{x - 1} = 4}$이고 $\displaystyle \lim\limits _{x \rightarrow 1} {\left( x - 1 \right) = 0}$이므로 $\displaystyle \lim\limits _{x \rightarrow 1} {\left\{ f \left( x \right) - a + 2 \right\} = 0}$ 함수 $f \left( x \right)$는 다항함수이므로 $\displaystyle \lim\limits _{x \rightarrow 1} {f \left( x \right)} = f \left( 1 \right) = a - 2$ $\displaystyle \lim\limits _{x \rightarrow 1} {\frac{f \left( x \right) - a + 2}{x - 1} = \lim\limits _{x \rightarrow 1} {\frac{f \left( x \right) - f \left( 1 \right)}{x - 1}} = f' \left( 1 \right)} = 4$ $\displaystyle \lim\limits _{x \rightarrow 1} {\frac{g \left( x \right) + a - 2}{x - 1}} = a$이고 $\displaystyle \lim\limits _{x \rightarrow 1} {\left( x - 1 \right)} = 0$이므로 $\displaystyle \lim\limits _{x \rightarrow 1} {\left\{ g \left( x \right) + a - 2 \right\}} = 0$ 함수 $g \left( x \right)$는 다항함수이므로 $\displaystyle \lim\limits _{x \rightarrow 1} {g \left( x \right) = g \left( 1 \right)} = - a + 2$ $\displaystyle \lim\limits _{x \rightarrow 1} {\frac{g \left( x \right) + a - 2}{x - 1}} = \lim\limits _{x \rightarrow 1} {\frac{g \left( x \right) - g \left( 1 \right)}{x - 1}} = g' \left( 1 \right) = a$ 함수 $f \left( x \right) g \left( x \right)$의 $x = 1$에서의 미분계수는 $f' \left( 1 \right) g \left( 1 \right) + f \left( 1 \right) g' \left( 1 \right)$$= 4 \times \left( - a + 2 \right) + \left( a - 2 \right) \times {a}$ $= a ^{2} - 6 a + 8$ $= - 1$ $a ^{2} - 6 a + 9 = 0$에서 $\left( a - 3 \right) ^{2} = 0$ 따라서 $a = 3$