2018년 03월 가형 21번

21. 함수 $f(x)=(x^2+ax+b)e^x$$f(x)=(x^2 +ax+b)e^x$과 함수 $g(x)$$g(x)$가 다음 조건을 만족시킨다.

(가) $f(1)=e,f^{\prime }(1)=e$$f(1)=e,~f'(1)=e$

(나) 모든 실수 $x$$x$에 대하여 $g(f(x))=f^{\prime }(x)$$g\big(f(x)\big)=f'(x)$이다.

함수 $h(x)=f^{-1}(x)g(x)$$h(x)=f^{-1}(x)g(x)$에 대하여 $h^{\prime }(e)$$h'(e)$의 값을 구하시오. (단, $a,b$$a,~b$는 상수이다.)

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i. 정리

뭐 딱히...

ii. 생각

• 우선 가능한 조건을 활용하자.

  $f(x)=(x^2+ax+b)e^x⟶f(1)=(1+a+b)e$$f(x)=(x^2 +ax+b)e^x\quad \longrightarrow \quad f(1)=(1+a+b)e$

  $f^{\prime }(x)=(x^2+ax+b+2x+a)e^2⟶f^{\prime }(1)=(1+a+b+2+a)e$$f'(x)=(x^2 +ax+b+2x+a)e^2\quad \longrightarrow \quad f'(1)=(1+a+b+2+a)e$

  정리하면,

  $\{\begin{array}{l}a+b=0\\ \\ 2a+b=-2\end{array}⟶a=-2,b=2$$\begin{cases} a+b=0\\ \\ 2a+b=-2\end{cases} \quad \longrightarrow \quad a=-2,~b=2$

  $\therefore f(x)=(x^2-2x+2)e^x,f^{\prime }(x)=x^2{e}^x$$\therefore~ f(x)=(x^2 -2x +2)e^x,~f'(x)=x^2 e^x$

• $h^{\prime }(x)$$h'(x)$를 구하자.

  보기 편하게 $j(x)=f^{-1}(x)$$j(x)=f^{-1}(x)$라 하자.

  $h(x)=j(x)g(x)⟶h^{\prime }(x)=j^{\prime }(x)g(x)+j(x)g^{\prime }(x)$$h(x)=j(x)g(x)\quad \longrightarrow \quad h'(x)=j'(x)g(x)+j(x)g'(x)$

  $h^{\prime }(e)=j^{\prime }(e)g(e)+j(e)g^{\prime }(e)$$h'(e)=j'(e)g(e)+j(e)g'(e)$

• $j(e),j^{\prime }(e)$$j(e),~j'(e)$를 구하자.

  $j(f(x))=x⟶j^{\prime }(f(x))f^{\prime }(x)=1$$j(f(x))=x\quad \longrightarrow \quad j'(f(x))f'(x)=1$

  $x=1$$x=1$을 대입하면 되겠다.

  $j(f(1))=j(e)=1$$j(f(1))=j(e)=1$

  $j^{\prime }(f(1))f^{\prime }(1)=j^{\prime }(e)e=1⟶\frac{1}{e}$$j'(f(1))f'(1)=j'(e)e=1\quad \longrightarrow \quad \cfrac 1e$

• $g(e),g^{\prime }(e)$$g(e),~g'(e)$를 구하자.

  $g(f(x))=f^{\prime }(x)$$g\big(f(x)\big)=f'(x)$

  $g^{\prime }(f(x))f^{\prime }(x)=f^{″}(x)$$g'\big(f(x)\big)f'(x)=f''(x)$

  여기도 $x=1$$x=1$을 대입하면 되네?

  $g(f(1))=g(e)=f^{\prime }(1)=e$$g\big(f(1)\big)=g\big(e\big)=f'(1)=e$

  $g^{\prime }(f(1))f^{\prime }(1)=g^{\prime }(e)e=f^{″}(1)⟶g^{\prime }(e)=\frac{f^{″}(1)}{e}$$g'(f(1))f'(1)=g'(e)e=f''(1)\quad \longrightarrow \quad g'(e)=\cfrac {f''(1)}e$

  $f^{″}(x)=(x^2+2x)e^x⟶f^{″}(1)=3e$$f''(x)=(x^2 +2x)e^x\quad \longrightarrow \quad f''(1)=3e$

$\therefore h^{\prime }(e)=\frac{1}{e}\cdot e+1\cdot 3=4$$\therefore~ h'(e)=\cfrac 1e \cdot e +1\cdot 3=4$