3.
a) f(x)=x³-3x²-9x+3 [-2;3]
f`(x)=3x²-6x-9=0
3x²-6x-9=0 |÷3
x²-2x-3=0 D=16
x₁=3 x₂=-1
f(-2)=(-2)³-3*(-2)²-9*(-2)+3=-8-12+18+3=1
f(-1)=(-1)³-3*(-1)²-9*(-1)+3=-1-3+9+3=8=fmax
f(3)=3³-3*3²-9*3=27-27-27+3=-24=fmin.
b) f(x)=2cosx-√2x [-π;π]
f`(x)=-2sinx-√2=0
-2sinx-√2=0 |÷(-2)
sinx=-√2/2
x₁=5π/4 x₂=-π/4
f(5/4)=2*cos(5π/4)-√2*5π/4=2*(-√2/2)-5√2π/4=-√2-5√2π/4=-√2*(1+5π/4)
f(-π/4)=2*cos(-π/4)-√2*(-π/4)=2*√2/2+√2π/4=√2+√2π/4=√2*(1+π/4)=fmax
f(-π)=2*cos(-π)-√2*(-π)=2*(-1)+√2*π=-2+√2π
f(π)=2*cosπ-√2*π=2*(-1)-√2*π=-(2+√2π)=fmin.
4.
y=2-x y=-x²+4
2-x=-x²+4
x²-x-2=0 D=9
x₁=2 x₂=-1
S=₋₁∫²(-x²+4-(2-x))dx=₋₁∫ ²(-x²+x+2)dx=(-x³/3+x²/2+2x) ₋₁|²=
=-2³/3+2²/2+2*2-((-1)³/3+(-1)²/2+2*(-1))=-8/3+2+4-(1/3+1/2-2)=
=3¹/₃+11¹/₆=10/3+7/6=27/6.
Ответ: S=27/6=4,5 кв.ед.