Lösung Tangente in einem Kurvenpunkt
Version 2.1 von Martin Stern am 2025/10/13 09:57
\(f(x)=\frac{1}{5}x^3-\frac{16}{5}x\)
\(f'(x)=\frac{3}{5}x^2-\frac{16}{5}\)
\(f'(3)=\frac{11}{5}\)
\(f(3)=-\frac{21}{5}\)
Einsetzen von \(m=\frac{11}{5}\) und \(P(3|-\frac{21}{5})\) in \(y=mx+c\):
\(-\frac{21}{5}=\frac{11}{5}\cdot 3+c\)
\(t: y= \frac{11}{5}x-\frac{54}{5}\)