+2점 · 주어진 함수를 미분하여 y ′ = e x x ( x + 1 ) ( x + 3 ) ( x + 6 ) \displaystyle y^\prime=e^xx(x+1)(x+3)(x+6) y ′ = e x x ( x + 1 ) ( x + 3 ) ( x + 6 ) (3-1) y ′ = e x x ( x + 1 ) ( x + 3 ) ( x + 6 ) \displaystyle y^\prime=e^xx(x+1)(x+3)(x+6) y ′ = e x x ( x + 1 ) ( x + 3 ) ( x + 6 )
+3점 · 도함수를 이용하여 함수의 증감을 판단하고 그래프의 개형을 그리면 따라서 그래프의 개형은 다음과 같다.
+4점 · 합성함수 미분법을 이용하여 g ( − 1 ) = 0 \displaystyle g(-1)=0 g ( − 1 ) = 0 g ( − 1 ) = − 2 a 3 \displaystyle g(-1)=-\frac{2a}3 g ( − 1 ) = − 3 2 a (3-2) 합성함수 미분법에 의해 f ′ ( g ( − 1 ) ) g ′ ( − 1 ) = 0 \displaystyle f^\prime(g(-1))g^\prime(-1)=0 f ′ ( g ( − 1 )) g ′ ( − 1 ) = 0 g ′ ( − 1 ) > 0 \displaystyle g^\prime(-1)>0 g ′ ( − 1 ) > 0 f ′ ( g ( − 1 ) ) = 0 \displaystyle f^\prime(g(-1))=0 f ′ ( g ( − 1 )) = 0 f ′ ( x ) = x ( 3 x + 2 a ) \displaystyle f^\prime(x)=x(3x+2a) f ′ ( x ) = x ( 3 x + 2 a ) g ( − 1 ) = 0 \displaystyle g(-1)=0 g ( − 1 ) = 0 g ( − 1 ) = − 2 a 3 \displaystyle g(-1)=-\frac{2a}3 g ( − 1 ) = − 3 2 a
+5점 · 함수의 성질로부터 g ( − 1 ) ≠ 0 \displaystyle g(-1)\ne0 g ( − 1 ) = 0 g ( − 1 ) = − 2 a 3 \displaystyle g(-1)=-\frac{2a}3 g ( − 1 ) = − 3 2 a +6점 · a = 3 e − 1 / 3 \displaystyle a=3e^{-1/3} a = 3 e − 1/3 하지만 f ( g ( − 1 ) ) = 4 e − 1 \displaystyle f(g(-1))=4e^{-1} f ( g ( − 1 )) = 4 e − 1 g ( − 1 ) ≠ 0 \displaystyle g(-1)\ne0 g ( − 1 ) = 0 g ( − 1 ) = − 2 a 3 \displaystyle g(-1)=-\frac{2a}3 g ( − 1 ) = − 3 2 a 4 27 a 3 = f ( g ( − 1 ) ) = 4 e − 1 \displaystyle \frac4{27}a^3=f(g(-1))=4e^{-1} 27 4 a 3 = f ( g ( − 1 )) = 4 e − 1 a = 3 e − 1 / 3 \displaystyle a=3e^{-1/3} a = 3 e − 1/3
+5점 · x \displaystyle x x g ( x ) \displaystyle g(x) g ( x ) (3-3) e a ( a + 3 ) 2 > a \displaystyle e^a(a+3)^2>a e a ( a + 3 ) 2 > a f ( g ( a ) ) = e a a 2 ( a + 3 ) 2 > a 3 > 4 a 3 27 \displaystyle f(g(a))=e^aa^2(a+3)^2>a^3>\frac{4a^3}{27} f ( g ( a )) = e a a 2 ( a + 3 ) 2 > a 3 > 27 4 a 3 이다. 함수 f \displaystyle f f 4 a 3 27 \displaystyle \frac{4a^3}{27} 27 4 a 3 g ( a ) \displaystyle g(a) g ( a ) g ( a ) > 0 \displaystyle g(a)>0 g ( a ) > 0
+2점 · g ( 0 ) = 0 \displaystyle g(0)=0 g ( 0 ) = 0 g ( 0 ) = − a \displaystyle g(0)=-a g ( 0 ) = − a +8점 · 사이값 정리를 적용하여 g ( 0 ) = 0 \displaystyle g(0)=0 g ( 0 ) = 0 또한 f ( g ( 0 ) ) = 0 \displaystyle f(g(0))=0 f ( g ( 0 )) = 0 g ( 0 ) = 0 \displaystyle g(0)=0 g ( 0 ) = 0 g ( 0 ) = − a \displaystyle g(0)=-a g ( 0 ) = − a g ( 0 ) = − a < 0 \displaystyle g(0)=-a<0 g ( 0 ) = − a < 0 g ( s ) = 0 \displaystyle g(s)=0 g ( s ) = 0 s \displaystyle s s 0 \displaystyle 0 0 a \displaystyle a a 0 = f ( g ( s ) ) = e s s 2 ( s + 3 ) 2 > 0 \displaystyle 0=f(g(s))=e^ss^2(s+3)^2>0 0 = f ( g ( s )) = e s s 2 ( s + 3 ) 2 > 0 g ( 0 ) = 0 \displaystyle g(0)=0 g ( 0 ) = 0
<참고>
문제의 미분가능한 함수 g \displaystyle g g f 1 , f 2 , f 3 \displaystyle f_1,f_2,f_3 f 1 , f 2 , f 3 f 1 ( x ) = x 2 ( x + 3 e − 1 / 3 ) ( x < − 2 e − 1 / 3 ) f 2 ( x ) = x 2 ( x + 3 e − 1 / 3 ) ( − 2 e − 1 / 3 ≤ x < 0 ) f 3 ( x ) = x 2 ( x + 3 e − 1 / 3 ) ( x ≥ 0 ) \displaystyle \begin{aligned}f_1(x)&=x^2(x+3e^{-1/3})\quad(x<-2e^{-1/3})\\f_2(x)&=x^2(x+3e^{-1/3})\quad(-2e^{-1/3}\le x<0)\\f_3(x)&=x^2(x+3e^{-1/3})\quad(x\ge0)\end{aligned} f 1 ( x ) f 2 ( x ) f 3 ( x ) = x 2 ( x + 3 e − 1/3 ) ( x < − 2 e − 1/3 ) = x 2 ( x + 3 e − 1/3 ) ( − 2 e − 1/3 ≤ x < 0 ) = x 2 ( x + 3 e − 1/3 ) ( x ≥ 0 )
함수 h ( x ) = e x x 2 ( x + 3 ) 2 \displaystyle h(x)=e^xx^2(x+3)^2 h ( x ) = e x x 2 ( x + 3 ) 2 g ( x ) = { f 1 ( h ( x ) ) ( x < − 1 ) f 2 − 1 ( h ( x ) ) ( − 1 ≤ x < 0 ) f 3 − 1 ( h ( x ) ) ( x ≥ 0 ) \displaystyle g(x)=\begin{cases}f_1(h(x))&(x<-1)\\f_2^{-1}(h(x))&(-1\le x<0)\\f_3^{-1}(h(x))&(x\ge0)\end{cases} g ( x ) = ⎩ ⎨ ⎧ f 1 ( h ( x )) f 2 − 1 ( h ( x )) f 3 − 1 ( h ( x )) ( x < − 1 ) ( − 1 ≤ x < 0 ) ( x ≥ 0 ) Correction. 참고의 첫 번째 식은 f 1 ( h ( x ) ) \displaystyle f_1(h(x)) f 1 ( h ( x )) f 1 − 1 ( h ( x ) ) \displaystyle f_1^{-1}(h(x)) f 1 − 1 ( h ( x )) f ( g ( x ) ) = h ( x ) \displaystyle f(g(x))=h(x) f ( g ( x )) = h ( x ) 로 두면 함수 g는 문제의 조건을 만족한다.
수리논술, 배움에서 논증의 완성까지.
