\(f(1) = 3 \text{,}\) dus \(A (1 , 3) \text{.}\)

\(f(x) = -4 x^{3} + 5 x^{2} + 4 x - 2\) geeft \(f'(x) = -12 x^{2} + 10 x + 4 \text{.}\)

Stel \(l{:}\,y = a x + b\) met \(a = f'(1) = 2 \text{.}\)

\(\begin{rcases}y = 2 x + b \\ \text{door } A (1 , 3)\end{rcases} \begin{matrix}2 ⋅ 1 + b = 3 \\ 2 + b = 3 \\ b = 1\end{matrix}\)
Dus \(l{:}\,y = 2 x + 1 \text{.}\)

\(f(4) = -5 \cdot 4^{2}+\frac{5}{2 \cdot 4-7} = -75 \text{,}\) dus \(A (4 , -75)\)

\(f(x) = -5 x^{2}+\frac{5}{2 x-7}\) geeft
\(f'(x) = -10 x+\frac{-10}{(2 x-7)^{2}}\)

\(l{:}\,y = a x + b\) met \(a = f'(4) = -10 \cdot 4+\frac{-10}{(2 \cdot 4-7)^{2}} = -50\)

\(\begin{rcases}y = -50 x + b \\ \text{door } A (4 , -75)\end{rcases} \begin{matrix}-50 ⋅ 4 + b = -75 \\ -200 + b = -75 \\ b = 125\end{matrix}\)
Dus \(l{:}\,y = -50 x + 125 \text{.}\)