Известно,что f(x)=0.5x^2-4 cравните: f(-5) и f(4) f(1) и f(-1) f(корень из 8) и f(- корень из 10)

Ответ:

GiViS

24.09.2020 17:59

$f(x)=0,5x^2-4\; \; \to \\\\F(x)=\int f(x)dx=\int (0,5x^2-4)dx=0,5\cdot \frac{x^3}{3}-4x+C\\\\1)\; \; F(-5)=0,5\cdot \frac{-125}{3}-20+C=-20\frac{5}{6}-20+C=-40\frac{5}{6}+C\\\\f(4)=0,5\cdot 4^2-4=8-4=4\\\\F(-5)\ \textgreater \ f(4),\; ecli\; \; -40\frac{5}{6}+C\ \textgreater \ 4\; \; \to \; \; C\ \textless \ 44\frac{5}{6}\\\\F(-5)=f(4)\; ,\; \; esli\; \; C=44\frac{5}{6} \\\\F(-5)\ \textless \ f(4)\; ,esli\; \; C\ \textgreater \ 44\frac{5}{6}$

$F(\sqrt8)=F(-\sqrt{10})\; ,\; \; esli\; \; C=1-16\sqrt2\\\\F(\sqrt8)\ \textless \ F(-\sqrt{10})\; ,\; \; esli\; \; C\ \textless \ 1-16\sqrt2$

$2)\; \; F(1)=\frac{1}{6}-4+C_1=-\frac{23}{6}+C_1\\\\F(-1)=-\frac{1}{6}+4+C_2=\frac{23}{6}+C_2\\\\F(1)\ \textgreater \ F(-1)\; ,esli\; \; -\frac{23}{6}+C_1\ \textgreater \ \frac{23}{6}+C_2\; \to \; \; C_1-C_2\ \textgreater \ \frac{46}{6}\; ,\\\\C_1\ \textgreater \ C_2+\frac{23}{3}\\\\F(1)=F(-1)=C_1=C_2+\frac{23}{3}\\\\F(1)\ \textless \ F(-1)\; ,\; esli\; \; C_1\ \textless \ C_2+\frac{23}{3}\\\\3)\; F(\sqrt8)=\frac{8\sqrt8}{6}-4\sqrt8+C=\frac{8}{3}\sqrt2-8\sqrt2+C=-16\sqrt2+C\\\\f(-\sqrt{10})=\frac{10}{2}-4=5-4=1\\\\F(\sqrt8)\ \textgreater \ F(-\sqrt{10})\; ,\; esli\; \; -16\sqrt2+C\ \textgreater \ 1\; \to \; \; C\ \textgreater \ 1-16\sqrt2$