Question

如何证明连续函数存在零点？

Answer 1

令f(x) = (1/2)^x-x^(1/3)
∵(1/2)^x 和 x^(1/3)都是连续函数
∴f(x) = (1/2)^x-x^(1/3)是连续函数
分别将x=0,1/6,1/3,1/2,1代入f(x) = (1/2)^x-x^(1/3)：
f(0)= (1/2)^0-0^(1/3)=1-0=1
f(1/6)= (1/2)^(1/6)-(1/6)^(1/3)=1/六次根号2 - 1/三次根号6=(三次根号6-六次根号2)/(六次根号2 *三次根号6)
∵三次根号6＞三次根号2＞六次根号2
∴f(1/6)＞0
f(1/3) = (1/2)^(1/3)-(1/3)^(1/3)=1/三次根号2 - 1/三次根号3=(三次根号3-三次根号2)/(三次根号2 *三次根号3)
∵三次根号3＞三次根号2
∴f(1/3)＞0
f(1/2) = (1/2)^(1/2)-(1/2)^(1/3)=1/根号2 - 1/三次根号2=(三次根号2-根号2)/(根号2 *三次根号2)
∵根号2＜三次根号2
∴f(1/2)＜0
f(1) = (1/2)^1-1^(1/3)=1/2-1=-1/2
因为f(1/3)*f(1/2)＜0
所以连续函数f(x)在x=1/3和x=1/2之间存在零点
即（1/2)的x次方=x的1/3次方有解x0,则x0在C.(1/3,1/2)