What is the principle of increasing and decreasing functions?

In the judgment of monotonicity of compound function f [g (x)], there is -- same increase different decrease
That is to say, if the monotonicity of F (x) and G (x) is the same in the same interval, then the composite function f [g (x)] is increasing in this interval
If their monotonicity is different, then the composite function f [g (x)] is decreasing in this interval
The monotonicity of F (x) = f (x) + G (x) is the same increase and decrease (in the same interval)

Proof of Euler's homogeneous function theorem (add 200 points)
Take the simplest binary function as an example
That is to say, when f (TX, ty) = T ^ NF (x, y)
There is FX (partial derivative) + FY = NF (x, y)
Sorry, it seems that the system only allows additional 100 points

You are wrong about this theorem
Should be:
When f (TX, ty) = T ^ NF (x, y), then x * FX + y * FY = NF (x, y)
It is proved that for f (TX, ty) = T ^ NF (x, y), the derivation of T on both sides is as follows:
x*f1(tx,ty)+y*f2(tx,ty)=n*t^(n-1)*f(x,y)
(where F1 is the derivative of the first variable)
Let t = 1, then we get x * FX + y * FY = NF (x, y)
The proof is complete
You can first increase the reward + 100 points, and then add 100 points when you adopt the answer

It involves a high number proof problem using zero point theorem,
Let f (x) be continuous on [a, b] and f (a) = f (b). It is proved that there exists XO belonging to (a, b) such that f (XO) = f (XO + (B-A) / 2)

Let f (x) = f (x) - f (x + (B-A) / 2), X belong to [a, (a + b) / 2], then f (a) + F ((a + b) / 2) = f (a) - f ((a + B) / 2) + F ((a + b) / 2) - f (b) = f (a) - f (b) = 0, so f (a) = f ((a + b) / 2) = 0 or one positive one negative 1, f (a) = f ((a + b) / 2) = 0, then take x0 = (a + b) / 2, obviously f (...)

The zero point theorem appears in the chapter of higher numbers
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In the first chapter, function and limit, section 10, the properties of continuous function on closed interval, the second point zero point theorem and intermediate value theorem appear! The fifth edition of Tongji's higher numbers!

In Lagrange's mean value theorem, why is continuity in a closed interval differentiable in an open interval? Can it be differentiable in a closed interval? Or are both closed intervals

If the open interval is continuous, f (a) and f (b) do not necessarily exist, and their existence does not necessarily conform to the theorem. You can design a function that monotonically increases in (a, b) but f (a) = f (b), which is open interval continuous, but the mean value theorem does not hold

I would like to ask the differential mean value theorem, why closed interval continuous, open interval differentiable?
Why does differential mean value theorem need closed interval continuous, open interval differentiable? Why is it not closed interval differentiable? That's easy to say, because differentiable must be continuous. Thank you

How can we say that it is differentiable in a closed interval? It is not differentiable if either the left limit does not exist or the right limit does not exist at the endpoint
But why is continuity a closed interval? Because continuity at the left end point means right continuity and vice versa
A function that is continuous at every point of an interval is called a continuous function on the interval. If the interval includes an end point, then the function that is continuous at the right end point is left continuous, and vice versa
Left continuous definition Lim X - > x0 - = f (x0 -) = f (x0)
For details, please refer to p61 of Tongji advanced mathematics fifth edition

Let f (x) be continuous on [a, b], differentiable on (a, b), and f (a) f (b) > 0, f (a) f ((a + b) / 2)

From F (a) f ((a + b) / 2) 0, we can see that there is x2 on ((a + b) / 2, b), such that f (x2) = 0, the constructor g (x) = f (x) / e ^ KX, G (x1) = g (x2) = 0, G (x) is derivable and continuous in [x1, X2], and there is at least one point ξ in (x1, x2), such that G '(ξ) = 0, that is, f' (ξ) = KF (ξ). In conclusion, for any real number k

Let f (x) be continuous on [1, e], 0

In the proof of mean value theorem of definite integral, it is proved that there is at least one point s in [a, b]. In this proof, the intermediate value theorem of continuous function is directly used, but shouldn't the intermediate value theorem of continuous function have at least one point s in (a, B)? It's a bit confusing

I know your doubts. Note that the intermediate value theorem considers two unequal function values (set as a and b), and can obtain any number between a and B (here is the open interval, because we consider the interval). Then look at its corollary, here is the closed interval, because what? Because here is the maximum and minimum, The maximum and minimum values must be obtained (there is no relationship between them here, a and B are taken into consideration). Let's take a look at the mean value theorem of definite integral. Here are also the maximum and minimum values, so it's right to use the closed interval

It is proved that the function y = PX ^ 2 + QX + R is always in the middle of the interval by using Lagrange mean value theorem

According to Lagrange's mean value theorem, in the interval [a, b], there exists a point ξ such that
y'(ξ) = (y(b)-y(a))/(b-a)
y'(ξ) = 2pξ+q
(y(b)-y(a))/(b-a) = p(a+b) +q
2pξ+q = p(a+b) +q
ξ = (a+b)/2

What is the principle of increasing and decreasing functions?