Question

Найдите точку минимума функции y=log5(x^2-30x+249)+8

Answer 1

$y=\log_5(x^2-30x+249)+8. \\ x^2-30x+249\ \textgreater \ 0, \\ D_{/4}=(-15)^2-249=225-249=-24\ \textless \ 0, \\ a=1\ \textgreater \ 0, \ D\ \textless \ 0, \ x^2-30x+249\ \textgreater \ 0 \ \forall x\in R, \\ D_y=R. \\ y'=\frac{1}{(x^2-30x+249)\ln5}\cdot(2x-30+0)+0=\frac{2x-30}{(x^2-30x+249)\ln5}. \\ \left [ {{2x-30=0,} \atop {x^2-30x+249=0,}} \right. \ \left [ {{x=15,} \atop {x\in \varnothing;}} \right. \\ x=15, \\ \begin{array}{c|ccc}x&(-\infty;15)&15&(15;+\infty)\\y'&-&0&+\\y&\searrow&\min&\nearrow\end{array} \\ x_{min}=15, \ y_{min}=\log_5(15^2-30\cdot15+249)+8=\log_524+8\approx10.$