Найдите наименьшее значение функции y=(x^2-4x+16)/x на отрезке [1;16]

[latex]y= \frac{x^2-4x+16}{x} =x-4+\frac{16}{x}\; ,\; \; ODZ:\; x\ne 0\\\\y'=1-\frac{16}{x^2}=\frac{x^2-16}{x^2}=0\; \; \to \; \; x^2-16=0\; ,\to \\\\x=\pm 4,\; x\ne 0\\\\+++(-4)---(0)---(4)+++\\\\\quad \nearrow \qquad (-4)\quad \searrow \quad (0)\quad \searrow (4)\quad \nearrow \\\\x_{max}=-4\; ,\; x_{min}=4\\\\x\in [\, 1,16\, ]\\\\y(1)=13\\\\y(4)=4\\\\y(16)=13\\\\y_{naimenshee}=y(4)=4[/latex]