<span>f(x)=(4x+1)/(x+3)
u=4x+1 u'=4 v=x+3 v'=1
</span>
f'(x)=1/v²[u'v-v'u]=1/(x+3)²[4<span>(x+3)-1*(4x+1)]=</span>11/(x+3)²
3x+12=0⇒3x=-12⇒x=-4
x²-2x=0
x(x-2)=0
x=0 U x=2
_ + _ +
----------------(-4)-------------(0)----------------(2)---------------------------
x∈(-∞;-4) U (0;2)
М и К принадлежат: у(-1,5)=-225, у(-3)=-900