Why does the original function not exist as long as there is a hop break?

1) Whether there is a jump break at x0 or not, it can be called a jump break point, and f(x) has a jump break point in the closed interval [a, b], which means that this break point should be a jump break point defined at x0.

2) f(x0) should exist (your piecewise function x=0 must have a value). If f(x0) does not exist, i.e. f(x) is not defined at x0, then x0 cannot be said to belong to (a, b).

3) If you ask for one'Original function', the'Original function'The derivative at the break point x0 does not exist, while f(x0) does.

That is, the original function f(x) of f(x) cannot be found such that f(x) derivative = f(x).

Personal insights.

The existence of a break point means that the function is discontinuous, and if it is not continuous, it cannot be integrated, so there is no original function.

Let's say you go somewhere in a foreign country, but your home is not next to an airport.

You may take a taxi, or take an airport bus, first to the airport, then take a flight, assuming you don't need a transfer, fly directly to an airport in the country, and then take a taxi or someone to pick you up, and finally to your destination.

In this process, there are domestic cars, planes, and foreign cars.

Can you say that you are coming to the country by country?

Can you tell that you flew directly to the country from your home?

Can you say that you are a foreign car to pick you up from home to the country?

To use a mathematical analogy, x>2 and x2 are two different things!

Although the integrand function is integrable, it does not indicate whether the integrative function is derivable, unless the integrand is continuous, i.e., f(x) continuous, then the integral must be derivable. If it is not continuous, the integral must not be derivable.

But as long as the integrand is integrable, the variable upper bound integral function must be continuous.

The integrable is a little less conditional than the original function to exist.

1.Definition of Functional Continuity:

Let the function f(x) be defined in a certain neighborhood of the point x0, and if lim(x x0)f(x)=f(x0), then f(x) is said to be continuous at the point x0.

If the function f(x) is continuous at every point in interval i, then f(x) is said to be continuous on interval i.

2.The function must meet three conditions at the same time for a function:

1) The function is defined at x0;

2) LIMF(X) exists at x-> x0;

3) When x-> x0, limf(x)=f(x0).

then the elementary function is continuous within its defined domain.

When the function has both left and right limits at a certain point, but is not equal, that point is the jump break point.

The variable upper bound integral function of a function with a hop break point is continuous. The variable upper bound integral function should appear something like |x|In this way, the segmented function is a case where the segmentation points are continuous, but not derivable.

Therefore, if there is a second type of discontinuity, such as infinite discontinuity, ** discontinuity, it is possible (but it is only possible, and it is not certain) that it cannot be accumulated. And if it is a finite first type (whether it is a jump break point or a departible break point), it must be integrable.

Sufficient conditions for the integrability of a function:

1. Theorem 1 lets f(x) be continuous over the interval [a,b], then f(x) is integrable over [a,b].

2. Theorem 2 assumes that f(x) is bounded in the interval [a,b] and there are only a finite number of first-class discontinuities, then f(x) is integrable on [a,b].

3. Theorem 3 lets f(x) be monotonically bounded on [a,b] between the districts and implicits, then f(x) is integrable on [a,b].

Integrable function is bounded.

Any integrable function must be bounded, but it is important to note that a bounded function is not necessarily integrable. A function that is not continuous at every point on its definition domain. The Dirichlet function is an example of a discontinuous function everywhere.

If f(x) is a function, and both the definition domain and the value range are real numbers, and if for every x, there is a >0 such that for every δ>0, y can be found, so that the following equation holds, then f(x) is a discontinuous function everywhere: 0< |x−y|<δ and|f(x)−f(y)|≥

The discontinuities that exist and are equal to the left and right limits are called discontinuities.

The unequal break point where the left and right limits exist is called the jump break.

The left and right limits are roughly matched with infinite discontinuities, which are called infinite discontinuities, in which infinity is an answer that can be solved, but it is generally regarded as the limit does not exist.

The discontinuity point where the left and right limit oscillations do not exist is called the oscillation discontinuity, where the oscillation is not the answer that cannot be solved, and the limit does not exist at all.

The answer is: false The definition of a continuous function is that the function y = f(x) is defined in a certain neighborhood of x point x0, and if lim(x->x0)f(x) = f(x0), then the function y = f(x) is said to be continuous at point x0. The definitions of left continuous and right continuous are similar to the above.

The discontinuity point of the function and its classification y = f(x) is said to be discontinuous at x0 if one of the following three conditions is not satisfied: (1) it is not defined at x0; (2) There is a definition but the limit does not exist (3) There is a definition, the limit exists, but it is not equal to f(x0) The discontinuous point x0 is called the discontinuity point or discontinuity point. The discontinuities that exist at both the left and right limits are called the first type, and the discontinuities that are not the first type are called the second type.

You can go to the break point, jump break point, infinite buried break point, and oscillation break point. The operation of continuous functions and the continuity of elementary functions (1) sum, difference, product, quotient continuum (2) composite functions and inverse functions properties of continuous functions on continuous intervals (1) boundedness and maximum minimum standby value theorem if function.

First of all, there is a theorem: if the function f is continuous on the interval i, then f has the original function f on i, i.e., f'(x) = f(x).

However, the function has only left and right succession at the jump break point and the go-break point, so there is no original function.

If the function f has only a finite number of first-class discontinuities on the interval [a,b], then f is said to be piecewise continuous on [a,b], and there are infinite discontinuities of discontinuity, so there is no original function.

You take a good look at the original function definition. Any x i has f (x) = f(x), then f is the original function of f. As long as f ≠ f, then f is not the original function of f.

This sentence should be said the other way around, it should be:

A function that is derivable over an interval and has no first-class discontinuities in that interval.

The above theorem (also known as the derivative continuity theorem) can be proved by the Lagrangian median theorem

If f(x) is in a neighborhood of x0 u(x0; δ), in the decentering neighborhood u°(x0; δ) and lim(x x0)f'(x) exists, then f(x) is also derivable at x0 and has f'(x0)=lim(x→x0)f'(x)

The first type of discontinuity point is defined as the function that exists at a certain point, but is not equal to the value of the function at that point.

Obviously, if the derivatives exist at a certain point and are equal to the left and right limits, then the derivatives are continuous at that point, and that point cannot be a discontinuity point.

And if the left and right limits of the derivative function exist at a certain point but are not equal, then the left limit of the derivative is the left derivative of the original function, and the right limit of the derivative is the right derivative of the original function. The unequal left and right limits mean that the left and right derivatives are unequal, so the original function is not derivative at that point, or the derivative function is not defined at that point. Therefore, the point will not be a jump break point (the definition of the first type of break point emphasizes that the point must have a function value, since there is no definition at the point, even if the left and right limits are not equal, it is not a jump break point).

In summary, the derivative function of a function that is derivable on a certain interval does not have a first-class discontinuity point in that interval.

If a function is interrupted at a certain point, it can be divided into the first type of breaks (the number of breaks and hop breaks) and the second type of breaks (infinite breaks and ** breaks).

If the function f(x) exists at x=x0 and the left limit f'(x0-0) is not equal to the right limit of f'(x0+0), x0 is called the hop break point of f(x); If the function f(x) exists at x=x0 and the left limit f'(x0-0) is equal to the f right limit'(x0+0), then x0 is said to be the devoidable discontinuity point of f(x).

To determine what type of discontinuity point a point belongs to, you need to find the left and right limits of the function at that point, and then judge them according to the definition.

I hope to help you, if you have any questions, please feel free to ask, I wish you progress in your studies!