Дано F(x)=x^3. Ее первообразная f(x)=Int (x^3)dx=1/4x^4+C. Если график проходит через точку М(1;-1), то -1=1/4*1^4+C. Откуда С=-5/4 и f(x)=1/4*x^4-5/4
[y=3-x
[y=2
3-X=2
-x=2-3
-x=-1
x=1
<span>1<=x^2<=1 [обл. опр. arccos] </span>
<span>x=[-1;1] </span>
<span>П/4-arccos(x^2)>=0 </span>
<span>arccos(x^2)<=П/4 </span>
<span>arccos(x^2)<=arccos(1/V2) [V-кв.корень] </span>
<span>arccos = убывающая ф-ция </span>
<span>x^2>=1/V2 </span>
<span>x=(-S;-1/2^(1/4)]U[1/2^(1/4);+S) </span>
<span>x=[-1;1] </span>
<span>=>x=[-1;-1/2^(1/4)]U[1/2^(1/4);1]</span>
Sin4x-sin(π/2-6x)=2*sin(4x-(π/2-6x))/2
*cos(4x+π/2-6x)/2=
2*sin(4x-π/2+6x)/2*cos((π/2-2x)/2=
2*sin(10x/2-π/2:2)*cos(π/2:2-2x:2)=
2*sin(5x-π/4)*cos(π/4-x)