Lösung Tangente in einem Kurvenpunkt

1.
\(f(x)=\frac{1}{5}x^3-\frac{16}{5}x\)
\(f'(x)=\frac{3}{5}x^2-\frac{16}{5}\)
\(f'(3)=\frac{11}{5}\)
\(f(3)=-\frac{21}{5}\)
Einsetzen von \(m=\frac{11}{5}\) und \(P(3|-\frac{21}{5})\) in \(y=mx+c\):
\(-\frac{21}{5}=\frac{11}{5}\cdot 3+c\)
\(t: y= \frac{11}{5}x-\frac{54}{5}\)

2.
Variante 1: \(f'(x)=m_t\)
\(\frac{3}{5}x^2-\frac{16}{5}=\frac{11}{5}\)
\(3x^2=27\)
\(x^2=9\)
\(x_1=3\) und \(x_2=-3\)
\(f(-3)=\frac{21}{5}=g(-3)\)

Variante 2: Argumentation mit Punktsymmetrie