We prove the claim by contradiction. Assume there is x(0) in (a, b] with f(x(0)) not equal to 0. We may assume that f(x(0)) > 0. Let A = { x : x in [a, x(0)] and f(x) = 0}. Since f(a) = 0, A is not empty. Further, it is bounded above so sup(A) exists. Let x(1) = sup(A). Thus there is a squence of numbers in A which converges to x(1). Since f is continuous, Limf(x(n)) = f(x(1)). And this means f(x(1)) = 0 since f(x(n)) = 0 for all n's. This shows x(1) < x(0). We also show that for all x's in (x(1), x(0)] we have that f(x) > 0. For if there is y in (x(1), x(0)] with f(y) < 0 or f(y) = 0, then if f(y) = 0, then y is in A and so y < x(1) or y = x(1), but y > x(1) a contradiction. If f(y) < 0, and coupled with f(x(0)) > 0 by Intermediate Value Theorem there is a z in (y, x(0)) such that f(z) = 0. so z is in A and then z < x(1), but z > y > x(1) meaning z > x(1) a contradiction again. Thus f(x) > 0 for all x's in (x(1), x(0)]. Next consider g(x) = ln(f(x)) on (x(1), x(0)]. g is differentiable on this interval. Let t be any number in (x(1), x(0)], and apply the mean value theorem for g on [t, x(0)] we have:
/g(t) - g(x(0))/ = /g'(c)//t - x(0)/ < M/t - x(0)/ < M*/x(1) - x(0)/.
Now let t--> x(1)+ we have /g(t) - g(x(0))/ ---> + infinity. A contradiction since it is bounded above by M*/x(1) - x(0)/.