If f(- 1)= 0, try to judge the number of zeros of the function f(x).

f(- 1)=a-b+c=0，b=a+c，b^2=( a+c)^2≥a^2+2ac+c^2≥4ac

f(x)=ax^2+(a+c)x+c=ax(x+ 1)+c(x+ 1)=(x+ 1)(ax+c)

Therefore, when a=c= 1, f(x) has two identical zeros-1, otherwise there are two different zeros, one is-1 and the other is-c/a.

(2) let f (x) = ax 2+bx+c, f (x1) = ax12+bx1+c = a, f (x2) = ax2 2+bx2+c = b,

The arithmetic mean of a and b (A+B)/2 is between a and b, and there must be x 1

The mean value theorem of continuous function is the content of higher mathematics. Try to give you a proof with elementary mathematics, from f (x1) = ax12+bx1+c = a, f (x2) = ax2 2+bx2+c = b,

f(x 1)-a =ax 1^2+bx 1+c-a=0，f(x2) -B =ax2^2+bx2+c-B=0，

That is, x 1 and x2 are the real roots of f (x)-a = 0 and f (x)-b = 0, respectively, which are obtained from the discriminant of quadratic equation.

b^2-4a(c-A)≥0，b^2-4a(c-B)≥0，

b^2≥4a(c-A)、b^2≥4a(c-B)，

The above two formulas are added together.

2b^2≥4a(c-A)+4a(c-B)=4a(2c-A-B)

b^2≥4a(c-(A+B)/2)

From the discriminant of quadratic equation, we know that the real root of f(x)-(A+B)/2=0, that is, x0 exists.

f(x0)=(A+B)/2，f(x0)= (f(x 1)+f(x2))/2

X 1

x 1=[-b √(b^2-4a(c-A))]/2

x2=[-b √(b^2-4a(c-B))]/2

x0=[-b √(b^2-4a(c-(A+B)/2))]/2

Since 4a(c-(A+B)/2 is between 4a (c-a) and 4a (c-b), no matter which root X 1 and X2 take, x0 must have a value between them.