Question

Помогите с 3 заданием,пожалуйста:(

Answer 1

$f(x)=\frac{x^{2}-8x}{x+1} \\\\1)f'(x)=\frac{(x^{2}-8x)#*(x+1)-(x^{2}-8x)*(x+1)'}{(x+1)^{2} }=\frac{(2x-8)(x+1)-(x^{2}-8x)*1 }{(x+1)^{2} }=\\\\=\frac{2x^{2}+2x-8x-8-x^{2}+8x}{(x+1)^{2} }=\frac{x^{2}+2x-8 }{(x+1)^{2} }\\\\\\2)f'(x)=0 \ \Rightarrow \ \frac{x^{2}+2x-8 }{(x+1)^{2} } =0\\\\x^{2}+2x-8=0\\\\(x+4)(x-2)=0\\\\x+4=0\\\\x_{1} =-4\\\\x-2=0\\\\x_{2}=2-neyd,tak \ kak \ 2\notin[-5;-2]$

$3)f(-5)=\frac{(-5)^{2}-8*(-5) }{-5+1}=\frac{25+40}{-4} =-16,25\\\\f(-4)=\frac{(-4)^{2}-8*(-4) }{-4+1}=\frac{16+32}{-3}=\boxed{-16-max}\\\\f(-2)=\frac{(-2)^{2}-8*(-2) }{-2+1}=\frac{4+16}{-1}=\boxed{-20-min}$

Answer 2

Ответ: f(наибол.)=f(-4)= -16 ,  f(наимен.)=f(-2)= -20  .

$f(x)=\dfrac{x^2-8x}{x+1}\ \ ,\ \ \ x\in [-5;-2\ ]\\\\\\f'(x)=\dfrac{(2x-8)(x+1)-(x^2-8x)\cdot 1}{(x+1)^2}=\dfrac{x^2+2x-8}{(x+1)^2}=0\ ,\ x\ne -1\ ,\\\\\\x^2+2x-8=0\ \ ,\ \ (x+1)^2-9=0\ \ ,\ \ (x+1-3)(x+1+3)=0\ ,\\\\(x-2)(x+4)=0\ \ ,\\\\x_1=2\notin [-5;-2\ ]\ \ ,\ \ \ x_2=-4\in [-5;-2\ ]\\\\f(-4)=\dfrac{16+32}{-3}=-\dfrac{48}{3}=-16\\\\f(-5)=\dfrac{25+40}{-4}=-\dfrac{65}{4}=-16\dfrac{1}{4}=-16,25\\\\f(-2)=\dfrac{4+16}{-1}=-20$

$\-20<-16,25<-16\ \ \Rightarrow \\\\f(naibol.)=f(-4)=-16\ \ ,\ \ f(naimen.)=f(-2)=-20$