\(f_t ( x ) =\frac{1}{16t}+ x^4 -x^2 + t + 6\) D\((-1|4,5)\)?
\(f_t ( -1 ) =\frac{1}{16t}+ 1-1 + t + 6=4,5\)
\(\frac{1}{16t}+ t =-\frac{3}{2} |\cdot16t\)
\(1+ 16t^2 =-24t \)
\( 16t^2+ 24t=-1 \)
\( t^2+ \frac{3}{2}t=-\frac{1}{16} \)
\( (t+ \frac{3}{4})^2=-\frac{1}{16}+ (\frac{3}{4})^2 =\frac{1}{2} |±\sqrt{~~}\)
1.)
\( t+ \frac{3}{4}=\frac{1}{2}\sqrt{2}\)
\( t_1 =\frac{1}{2}\sqrt{2}-\frac{3}{4}\)
2.)
\( t+ \frac{3}{4}=-\frac{1}{2}\sqrt{2}\)
\( t_2 =-\frac{1}{2}\sqrt{2}-\frac{3}{4}\)
Wird fortgesetzt.