\(f(2) = 6 \text{,}\) dus \(A (2 , 6) \text{.}\)

\(f(x) = -x^{3} + 4 x^{2} - 2 x + 2\) geeft \(f'(x) = -3 x^{2} + 8 x - 2 \text{.}\)

Stel \(l{:}\,y = a x + b\) met \(a = f'(2) = 2 \text{.}\)

\(\begin{rcases}y = 2 x + b \\ \text{door } A (2 , 6)\end{rcases} \begin{matrix}2 ⋅ 2 + b = 6 \\ 4 + b = 6 \\ b = 2\end{matrix}\)
Dus \(l{:}\,y = 2 x + 2 \text{.}\)

\(f(3) = 3 \cdot 3+\frac{5}{-3 \cdot 3+8} = 4 \text{,}\) dus \(A (3 , 4)\)

\(f(x) = 3 x+\frac{5}{-3 x+8}\) geeft
\(f'(x) = 3+\frac{15}{(-3 x+8)^{2}}\)

\(l{:}\,y = a x + b\) met \(a = f'(3) = 3+\frac{15}{(-3 \cdot 3+8)^{2}} = 18\)

\(\begin{rcases}y = 18 x + b \\ \text{door } A (3 , 4)\end{rcases} \begin{matrix}18 ⋅ 3 + b = 4 \\ 54 + b = 4 \\ b = -50\end{matrix}\)
Dus \(l{:}\,y = 18 x - 50 \text{.}\)