f(x)=?

GRF 3. Grades: f(x)

H(-1|8)

Gerade: g(x)=-4x+4 berührt f(x) bei x=1

$$\small{\text{$ \begin{array}{l|rclr} \mathbf{ \mathrm{GRF~}3.\mathrm{Grades:~} } & \mathbf{f(x)} & \mathbf{=} & \mathbf{ ax^3+bx^2+cx+d }\\ & \mathbf{f'(x)} & \mathbf{=} & \mathbf{ 3ax^2+2bx+c }\\ \\ \hline \\ \mathrm{H(-1|8)} \quad f(-1)=8 & a(-1)^3+b(-1)^2+c(-1)+d &=& 8\\ & -a+b-c+d &=& 8 & (1)\\ &\\ \mathrm{H(-1|8)} \quad f'(-1)=0 & 3a(-1)^2+2b(-1)+c &=& 0\\ & 3a-2b+c &=& 0 & (2)\\ \\ \hline \\ g(1)=0 \quad f(1)=0 & a(1)^3+b(1)^2+c(1) +d &=& 0\\ & a+b+c+d &=& 0 & (3)\\ &\\ g'(1)=-4 \quad f'(1)=-4 & 3a(1)^2+2b(1)+c &=& -4\\ & 3a+2b+c &=& -4 & (4)\\ \\ \hline \\ (1) & -a+b-c+d &=& 8 \\ (2) & 3a-2b+c &=& 0 \\ (3) & a+b+c+d &=& 0 \\ (4) & 3a+2b+c &=& -4 \\ \\ \hline \\ (4)-(2) & 3a-3a+2b+2b+c-c &=& -4\\ & 4b &=& - 4 \\ & \mathbf{b} & \mathbf{=} & \mathbf{-1} \\ \\ \hline \\ (3)+(1) & a-a+b+b+c-c+d+d &=& 8\\ & 2b+2d &=& 8 \quad | \quad :2 \\ & b+d &=& 4 \\ & (-1)+d &=& 4 \\ & d &=& 4+1 \\ & \mathbf{d} & \mathbf{=} & \mathbf{5} \\ \\ \hline \\ (2)-(3) & 3a-a-2b-b+c-c-d &=& 0\\ & 2a-3b-d &=& 0 \\ & 2a-3(-1)-5 &=& 0 \\ & 2a+3-5 &=& 0 \\ & 2a-2 &=& 0 \\ & 2a &=& 2 \quad | \quad :2 \\ & \mathbf{a} & \mathbf{=} & \mathbf{1} \\ \\ \hline \\ (3) & c &=& -a-b-d\\ & c &=& -1-(-1)-5 \\ & c &=& -1+1-5\\ & \mathbf{c} & \mathbf{=} & \mathbf{-5} \\ \end{array} $}}$$

$$\mathbf{ \mathrm{GRF~}3.\mathrm{Grades:~} } \mathbf{f(x)= x^3-x^2-5x+5 }$$