Question

Задания на скриншоте! СРОЧНООО!

Answer 1

$1)\; \; \; y=2\sqrt{x}+x^2\; \; ,\; \; x_0=1\\\\y'=\dfrac{1}{\sqrt{x}}+2x=\dfrac{1+2x\sqrt{x}}{\sqrt{x}}\; ,\\\\y'(1)=\dfrac{1+2}{1}=3\; \; ,\; \; \; y(1)=2+1=3\; ,\\\\y=y(x_0)+y'(x_0)(x-x_0)\\\\y=3+3(x-1)\; \; \Rightarrow \; \; \; \underline {\; y=3x+2\; }$

$2)\; \; y=\dfrac{x^2+4}{2x-3}\; \; ,\; \; x\in [\, 0;5\; ]\\\\\\y'=\dfrac{2x(2x-3)-2(x^2+4)}{(2x-3)^2}=\dfrac{2x^2-6x-8}{(2x-3)^2}=\dfrac{2\, (x^2-3x-4)}{(2x-3)^2}=0\\\\\\x^2-3x-4=0\; \; \to \; \; \; x_1=-1\; ,\; x_2=4\; \; (teorema\; Vieta)\\\\x=-1\notin [\, 0;5\, ]\\\\y(0)=-\dfrac{4}{3}=-1\dfrac{1}{3}\\\\y(4)=\dfrac{16+4}{8-3}=\dfrac{20}{5}=4\\\\y(5)=\dfrac{25+4}{10-3}=\dfrac{29}{7}=4\dfrac{1}{7}\\\\\\y_{naimen.}=y(0)=-1\dfrac{1}{3}\; \; ,\; \; \; y_{naibol.}=y(5)=4\frac{1}{7}$