\(f(1) = 2 \text{,}\) dus \(A (1 , 2) \text{.}\)

\(f(x) = -4 x^{3} + 4 x^{2} + 5 x - 3\) geeft \(f'(x) = -12 x^{2} + 8 x + 5 \text{.}\)

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

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

\(f(1) = 5 \cdot 1^{3}+(4 \cdot 1-2)^{3} = 13 \text{,}\) dus \(A (1 , 13)\)

\(f(x) = 5 x^{3}+(4 x-2)^{3}\) geeft
\(f'(x) = 15 x^{2}+12 (4 x-2)^{2}\)

\(l{:}\,y = a x + b\) met \(a = f'(1) = 15 \cdot 1^{2}+12 (4 \cdot 1-2)^{2} = 63\)

\(\begin{rcases}y = 63 x + b \\ \text{door } A (1 , 13)\end{rcases} \begin{matrix}63 ⋅ 1 + b = 13 \\ 63 + b = 13 \\ b = -50\end{matrix}\)
Dus \(l{:}\,y = 63 x - 50 \text{.}\)