Zero theorem: Let f(x) be continuous on the closed interval [a, b] (the so-called continuity can be understood as a function like a continuous curve), and f (a) f (b).

This problem can be proved by constructing auxiliary function and using zero theorem.

It is proved that if the constructor g(x) = f (x)-[f (x1)+f (x2)]/2, then g (x) is continuous on r and ∴ is continuous on [x 1, x2].

g(x 1)= f(x 1)-[f(x 1)+f(x2)]/2 =[f(x 1)-f(x2)]/2

g(x2)= f(x2)-[f(x 1)+f(x2)]/2 =[f(x2)-f(x 1)]/2

Since the condition f(x 1)≠f(x2) is known, f(x 1)-f(x2) and f(x2)-f(x 1) have different signs.

That is, g(x 1)g(x2)=0.

∴ There is at least one x0∈(x 1, x2), so g (x0) = f (x0)-[f (x1)+f (x2)]/2 = 0.

That is f(x0)=[f(x 1)+f(x2)]/2.