Since f'(x) has the same sign at a and b, it is not necessary to assume that f'(x) is positive at a and b.

If f'(a)>0 and f '(x) is continuous, there is a right neighborhood of a, so that in this neighborhood, f' (x) >: 0,

That is, in this neighborhood, the function monotonically increases, so there is C >；; A, make f (c) >; f(a)=0

Similarly, if f'(b)>0 and f '(x) is continuous, there is a left neighborhood of b, so that in this neighborhood, f' (x) >: 0,

That is to say, in this neighborhood, the function monotonically increases, so there is D.

Therefore, if the signs of f(x) at c and d are different, then k∈(a, b) exists, so f(k)=0.

Therefore, f(x) has at least three zeros.

2. The constructor G (x) = e xf (x) is first-order derivable in [a, b] and second-order derivable in [a, b].

g(a)=g(k)=g(b)=0

According to Rolle's theorem, there are k 1∈(a, k) and k2∈(k, b) such that: g'(k 1)=g'(k2)=0.

According to Rolle, there exists t∈(k 1, k2), so: g''(t)=0.

g'(x)=e^xf '(x)+e^xf(x)

g''(x)=e^xf ''(x)+2e^xf '(x)+e^xf(x)

Therefore: e t f'' (t)+2e t f' (t)+e t f (t) = 0.

The conclusion is: f ''(t)+2f '(t)+f(t)=0.

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