Question

Задание В15 0,5х^2 -17х +70lnх -7 найдите точку минимума

Answer 1

$F(x)=0.5x^2-17x+70lnx-7 \\ x\ \textgreater \ 0 \\ F'(x)=x-17+ \frac{70}{x} \\ F'(x)=0;x-17+ \frac{70}{x} =0 |*x\\ x^2-17x+70=0 \\ D=289-280=9;a=1;b=-17;c=70;D=b^2-4ac \\ x_1= \frac{17-3}{2}= \frac{14}{2}=7 \\ x_2= \frac{17+3}{2}= \frac{20}{2}=10 \\ F'(x)\ \textgreater \ 0; 0\ \textless \ x\ \textless \ 4;x\ \textgreater \ 13 \\ F'(x)\ \textless \ 0; \\ 4\ \textless \ x\ \textless \ 13 \\ x=13;y_{min} =0.5(13)^2-17*13+70ln13-7= \frac{169}{2}-221+70ln13- \\ -7= \frac{169-442-14}{2}+70ln13=- \frac{287}{2}+70ln13$