Let me give you one last performance.

The topic is to study the function y=ax? -the properties of bx, a and b are unchanged, and a is not zero. For x, when the values are-1, -0.72, -0.44, -0. 16, 0. 12, 0.4, the corresponding values of y are 4, 1.25, 0.02.

1 Whether the function f(x) has zero in the interval (0.4, 0.44), write out the judgment and prove it.

It is proved that the function f(x) monotonically decreases in the interval (negative infinity, -0.3).

It has been found that A(- 1, 4), B(t, f (t)) (- 1

The first problem is existence.

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