Question

помогите с 10 и 12 пж

Answer 1

Ответ:

$10)\ \ y=4x+\dfrac{1}{x}+7\ \ ,\ \ x\in (\ 0\ ;\ 2\ ]\\\\y'=4-\dfrac{1}{x^2}=\dfrac{4x^2-1}{x^2}=\dfrac{(2x-1)(2x+1)}{x^2}=0\ \ ,\ \ x_1=-\dfrac{1}{2}\ ,\ x_2=\dfrac{1}{2}\\\\znaki\ y'(x):\ \ +++[-\frac{1}{2}\, ]---(0)---[\, \frac{1}{2}\, ]+++\\{}\qquad \qquad \qquad \quad \nearrow \quad max\ \ \ \searrow \qquad \quad \searrow \quad min\ \nearrow \\\\\\x=\dfrac{1}{2}\in (\ 0\ ;\ 2\ ]\ \ ,\ \ y(\frac{1}{2})=2+2+7=11\\\\x=2\in (\ 0\ ;\ 2\ ]\ \ ,\ \ y(2)=8+0,5+7=\boxed{15,5}\ \ -\ \ naibolshee$

$12)\ \ f(x)=-12\, ln(x^2-12)+6x\\\\f'(x)=-12\cdot \dfrac{2x}{x^2-12}+6=\dfrac{-24x+6x^2-72}{x^2-12}=0\ ,\ \ x\ne \pm 2\sqrt3\ ,\\\\\\6x^2-24x-72=0\ \ ,\ \ x^2-4x-12=0\ \ ,\ \ x_1=-2\ ,\ x_2=6\\\\znaki\ f'(x):\ \ +++(-2\sqrt3)+++[\, -2\ ]---(2\sqrt3)---[\ 6\ ]+++\\{}\qquad \qquad \qquad \quad \nearrow\qquad \qquad \qquad \nearrow \quad max\quad \searrow\qquad \qquad \ \ \searrow\ \ min\ \ \nearrow\\\\\\\boxed{x(min)=6}\\\\y(min)=-12\cdot ln(36-12)+6\cdot 6=-12\cdot ln24+36\approx -2,1366$