Question

Найдите точки экстриума функции у=(х+3)²(х-5)
Даю 40баллов

Answer 1

$y=(x+3)^2(x-5)\\\\y'=2(x+3)(x-5)+(x+3)^2\cdot 1=(x+3)(2x-10+x+3)=\\\\=(x+3)(3x-7)=0\; \; \to \; \; \; x_1=-3\; ,\; x_2=\frac{7}{3}\\\\znaki\; y':\; \; \; +++(-3)---(\frac{7}{3})+++\\\\.\qquad \qquad \quad \; \; \nearrow \; \; \; (-3)\; \; \searrow \; \; \; (\frac{7}{3})\; \; \nearrow \\\\x_{max}=-3\; \; ,\; \; \; x_{min}=\frac{7}{3}\\\\y_{max}=y(-3)=(-3+3)^2(-3-5)=0\\\\y_{min}=y(\frac{7}{3})=(\frac{7}{3}+3)^2(\frac{7}{3}-5)=(\frac{16}{3})^2\cdot \frac{-8}{3}=-\frac{2048}{27}=-75\frac{23}{27}$