By the IVT, show that p(x)=x3+x−1p(x) = x^3 + x - 1p(x)=x3+x−1 has a root in (0,1)(0,1)(0,1).
p(0)=1>0p(0)=1>0p(0)=1>0 and p(1)=1>0p(1)=1>0p(1)=1>0, so no root in (0,1)(0,1)(0,1)
p(0)=−1<0p(0)=-1<0p(0)=−1<0 and p(1)=1>0p(1)=1>0p(1)=1>0, so by IVT there exists c∈(0,1)c\in(0,1)c∈(0,1) with p(c)=0p(c)=0p(c)=0
p(0)=0p(0)=0p(0)=0, so x=0x=0x=0 is already a root
IVT does not apply to cubics