Because f(-x)=-f(x) is odd function, so f(0.44)=-f(-0.44)=-0.02 plus or minus 0.0 1: 0.

Therefore, there must be a value between (0.4, 0.44) to make f(x)=0.

The second question:

Let x1> If x2 belongs to (negative infinity, -0.3), then f (x1)-f (x2) = (x1-x2) [a (x1+x2/2) 2+(3a/4) x22+b].

According to the data in the table, f (-1) =-(a+b) = 4 >; 0，f(0.4)= 0.064 a+0.4b = 0.8 & gt； 0. get an a

And the maximum value l (max) = 0.63a+b.

Let the function g (x) = ax 2+b, then f(x)=xg(x) is between x=0 and x belongs to (-0.44, -0.4) and (0.4, 0.44), and f(x)=0.

So g (0.5) = 0.25a+b.

l(max)= 0.63 a+b & lt； g(0.5)& lt； 0 so x1>; F (x 1) < F (x2) is a constant, which proves that f (x) decreases at (negative infinity, -0.3).

Three questions:

The conclusion is correct. In fact, this conclusion is called Lagrange mean value theorem in higher mathematics.

It is proved that the auxiliary function f (x) = f (x)-f (-1)-[f (t)-f (-1)] * (x+1).

It is easy to verify that F(t)=F(- 1)=0, and f (x)' = f (x)'-(f (t)-f (-1))/(t+1) can be used in (-6544).

It is proved that f (m)' = (f (t)-f (-1))/(t+1).

Let f (x)'-(f (t)-f (-1))/(t+1) = 0 because b-(f (t)-f (-1))/(t+1. 0 is constant.

Find the root of the equation x =+-√((t- 1/2)2+3/4))/3.

So when the minimum value of x is t=2, x=- 1.

Make x < T >；; 1/2 or t

So- 1