f(x)=?

GRF 3. Grades: f(x)

H(-1|8)

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

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

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