∵f(x)=(3x-1)(x³-7x²-2x+1)=0⇒x=\frac{1}{3}

設g(x)=x³-7x²-2x+1⇒g(0)·g(1)=1×(-7)<0

∴f(x)=0在0與1之間恰有一無理根

∵f(0)=-1

n = 2 (slider from 0 to 5, step 1)
f(x) = 3 * x^(4) - 22 * x^(3) + x^(2) + 5 * x - 1