ни четная ни нечетная;
0, x\in(0;+\infty) \ \ y>0, \\
x<0, x\in(-\infty;-1)\cup(1;0) \ \ y<0; \\
5) f'(x)=(\frac{4x}{(x+1)^2})'= \frac{4x'(x+1)^2-4x((x+1)^2)'}{(x+1)^4}=\frac{4(x+1)^2-8x(x+1)(x+1)'}{(x+1)^4}=\\=\frac{4(x+1)^2-8x(x+1)}{(x+1)^4}=\frac{4(x+1)(x+1-2x)}{(x+1)^4}=\frac{4(1-x)}{(x+1)^3}; \\
x+1 \neq 0, x \neq -1; \\
f'(x)=0, \frac{4(1-x)}{(x+1)^3}=0, x=1, " alt="4) y\gtrless0, \frac{4x}{(x+1)^2}\gtrless0, \\ x(x+1)^2\gtrless0, \\
x>0, x\in(0;+\infty) \ \ y>0, \\
x<0, x\in(-\infty;-1)\cup(1;0) \ \ y<0; \\
5) f'(x)=(\frac{4x}{(x+1)^2})'= \frac{4x'(x+1)^2-4x((x+1)^2)'}{(x+1)^4}=\frac{4(x+1)^2-8x(x+1)(x+1)'}{(x+1)^4}=\\=\frac{4(x+1)^2-8x(x+1)}{(x+1)^4}=\frac{4(x+1)(x+1-2x)}{(x+1)^4}=\frac{4(1-x)}{(x+1)^3}; \\
x+1 \neq 0, x \neq -1; \\
f'(x)=0, \frac{4(1-x)}{(x+1)^3}=0, x=1, " align="absmiddle" class="latex-formula">
x=-1 - точка разрыва,
х=1 - критическая точка,
0, y\nearrow, \\ x>1, x\in(1;\infty), y'<0, y\searrow," alt="6) f'(x)\gtrless0, \frac{4(1-x)}{(x+1)^3}\gtrless0, \\ (1-x)(x+1)^3\gtrless0, \\ (x-1)(x+1)^3\lessgtr0, \\ x<-1, x\in(-\infty;-1), y'<0, y\searrow, \\ -1<x<1, x\in(-1;1), y'>0, y\nearrow, \\ x>1, x\in(1;\infty), y'<0, y\searrow," align="absmiddle" class="latex-formula">
x=1 - точка максимума,
2, y''>0, y\smallsmile," alt="f''(x)= (\frac{4(1-x)}{(x+1)^3})'=4 \frac{(1-x)'(x+1)^3-(1-x)((x+1)^3)'}{(x+1)^6}=\\=4 \frac{-(x+1)^3-3(1-x)(x+1)^2(x+1)'}{(x+1)^6}=-4 \frac{(x+1)^3+3(1-x)(x+1)^2}{(x+1)^6} =\\=\frac{-4(x+1)^2(x+1+3-3x)}{(x+1)^6}=\frac{-4(4-2x)}{(x+1)^4}=\frac{8(x-2)}{(x+1)^4}, \\f''(x)\gtrless0, \frac{8(x-2)}{(x+1)^4}\gtrless0, \\ (x-2)(x+1)^4\gtrless0, \\ x<-1, y''<0, y\smallfrown, \\ -1<x<2, y''<0, y\smallfrown, \\ x>2, y''>0, y\smallsmile," align="absmiddle" class="latex-formula">
x=2 - точка перегиба.
