
1. \(f(x)=y=x^2\) \(f\ '(x)=y \ '=2x\)
2. \(g(x)= mx+1\) \(m_{g(x)}=-\frac{1}{f\ '(x)}\)
\(d^2=D=x_Q^2+(1-y_Q)^2\)
\(y_Q=x_Q2\)
\(D=x_Q^2+(1-x_Q^2)^2\)
\(D=x_Q^2+1-2x_Q^2+x_Q^4\) \([x_Q=x]\)
\(h(x)=D=1-x^2+x^4\)
\(h\ '(x)=4x^3-2x=0\)
\(x(4x^3-2)=0\) x1 = 0 entfällt
\(x^3=\frac{1}{2}\) \([x=x_Q]\)
\(x_Q=\frac{1}{\sqrt[3]{2} }=0,793700525987\) \(y_Q=x_Q^2=\frac{1}{ \sqrt[3]{4} }=0,629960524947\)
\(D=x_Q^2+(1-y_Q)^2\)
\( D=x_Q^2+(1-x_Q^2)^2\)
\(D=(\frac{1}{\sqrt[3]{2}})^2+(1-\frac{1}{\sqrt[3]{4}})^2=0.766889738049\)
\(d=+\sqrt D=0,875722409242\)
Danke heureka
!