If the function is continuous at a,b, then it must be unilaterally derivable at each point of a,b, r

Right. According to, continuous and left (right) continuous definitions:

and left (right) derivative definitions:

It can be seen that this proposition is correct, and the proof is given below.

Let f(x) be continuous in (a,b), and 0 is an arbitrarily small positive number, and take any point x0(a,b), then f(x) is continuous to the left of x0, lim(x->x0-)=f(x0)

Defined by the derivative, y'=lim(x->x0-) f(x)-f(x0)) (x-x0) exists, i.e., from continuous = left continuous = left derivable, and from continuous deducible right continuous, in the same way, f(x) is right-derivable at x0, so f(x) is unilateral derivable at (a, b) at each point.

I think you should have both continuous and derivable definitions, I won't repeat them, the proof process is not perfect, you can improve it yourself, and if you have problems in the future, think about it yourself, and then ask others if you really can't think of it, that's it.

False, counterexample: f(x)=xsin(1 x) x is not equal to 0

0 x = 0 Obviously, the function is continuous (and consistently continuous) at (-1,1).

But defined by the derivative, f'(x)=lim(x->0+(-f(x)-f(0)) (x-0)=lim(x->0+(-xsin(1 x) x=lim(x->0+(-sin(1 x) does not exist.

False, counterexample: f(x)=xsin(1 x) x is not equal to 0

0 x=0 is obviously pretending to be a servant, and the function is laughing (-1,1) continuous (and consistently).

But the indiscretion is defined by the derivative, f'(x)=lim(x->0+(-f(x)-f(0)) (x-0)=lim(x->0+(-xsin(1 x) x=lim(x->0+(-sin(1 x) does not exist.

Since f(x) is continuous in the closed interval, the open interval can be induced, and ab>0 This function must have a little answer in the open interval a,b (a,b).

Let g(x) =x 2 be continuous on [a,b] and be derivable within (a,b).

Then Cauchy's median theorem: (f(b)-f(a)) g(b)-g(a))=f'(ξg'(ξ

So 2 [f(b)-f(a)]=b 2-a 2)f'(ξ

WILL BE ... Consider the slope.

This is actually the Lagrangian median theorem.

Is this simple?,High school side closure problem.。

First of all, the monotonous old function, there is only an increase or a decrease function.

If 2 points, x1, x2, x1>a, and x20 are set on the interval (a,b), then it is an increasing function, the minimum value is f(a), and the maximum value is f(b);

2: f(x1)-f(x2)<0, then it is a subtraction function, the maximum value is f(a), and the minimum value is f(b);

3: (x1)-f(x2)=0, then it is a straight line parallel to the x-axis housing crack< b, there is f(x1)-f(x2) compared to 0.

Hello function f(x) is continuous on [a,b], derivable within (a,b), ad f(x2)-f(x1)=f'(e) (x2-x1) e belongs to (x1, x2).