자연계열 3번

3-1.[풀이]$\displaystyle y=(x-4)e^x\Rightarrow y^{\prime}=(x-3)e^x$곡선 위의 점 $\displaystyle (x_1,(x_1-4)e^{x_1})$에서의 접선의 방정식 $\displaystyle y-(x_1-4)e^{x_1}=(x_1-3)e^{x_1}(x-x_1)$$\displaystyle (b,0)$을 지나므로 $\displaystyle -(x_1-4)e^{x_1}=(x_1-3)e^{x_1}(b-x_1)$$\displaystyle \Rightarrow-(x_1-4)=(x_1-3)(b-x_1)$$\displaystyle \Rightarrow(x_1-b)(x_1-3)=x_1-4$$\displaystyle \Rightarrow x_1^2-(3+b)x_1+3b=x_1-4$$\displaystyle \Rightarrow x_1^2-(4+b)x_1+3b+4=0$서로 다른 두 실근 : $\displaystyle (4+b)^2-4(3b+4)>0$$\displaystyle \Rightarrow b^2+8b+16-12b-16>0$$\displaystyle \Rightarrow b^2-4b>0$$\displaystyle \Rightarrow b>4,b<0$정수 : $\displaystyle b=0,1,2,3,4$

3-2.[풀이]$\displaystyle f^{\prime}(g(t))=t$$\displaystyle g(t)=x$라하면 $\displaystyle t=f^{\prime}(g(t))=f^{\prime}(x)=(x-3)e^x$$\displaystyle \Rightarrow g((x-3)e^x)=x$양변을 미분하면$\displaystyle \Rightarrow g^{\prime}((x-3)e^x)\cdot(x-2)e^x=1$$\displaystyle \Rightarrow g^{\prime}((x-3)e^x)=\frac1{(x-2)e^x}$$\displaystyle g^{\prime}(e^4)=\frac1{2e^4}$