Can you specify how to prove that a function is continuous and differentiable in a certain (open and closed) interval? I already know the continuity and differentiability at a certain point! Search a lot of answers, feel only said in a certain point, or did not say clearly! Like the following topic! In the first step, how do you know that it is differentiable and continuous in the interval of real number axis? It is not necessary to find out the derivative of the function f (x) = (x-1) (X-2) (x-3) (x-4) (X-5). It is shown that the equation F & # 61602; (x) = 0 has several real roots, and the interval in which they are located is pointed out. Because the function f (x) = (x-1) (X-2) (x-3) (x-4) (X-5) is continuous and differentiable on the whole real axis, and f (1) = f (2) = f (3) = f (4) = f (5) = 0, applying Rolle's theorem in the interval (1,2), (2,3), (3,4), (4,5) respectively, we can obtain that the equation F &; (x) = 0 has at least four real roots, but because F &; (x) is a polynomial of degree 4, there are at most four real roots, so the equation F &; (x) has four real roots; (x) = 0 has only four real roots and is located in the interval (1,2), (2,3), (3,4), (4,5) respectively

Can you specify how to prove that a function is continuous and differentiable in a certain (open and closed) interval? I already know the continuity and differentiability at a certain point! Search a lot of answers, feel only said in a certain point, or did not say clearly! Like the following topic! In the first step, how do you know that it is differentiable and continuous in the interval of real number axis? It is not necessary to find out the derivative of the function f (x) = (x-1) (X-2) (x-3) (x-4) (X-5). It is shown that the equation F & # 61602; (x) = 0 has several real roots, and the interval in which they are located is pointed out. Because the function f (x) = (x-1) (X-2) (x-3) (x-4) (X-5) is continuous and differentiable on the whole real axis, and f (1) = f (2) = f (3) = f (4) = f (5) = 0, applying Rolle's theorem in the interval (1,2), (2,3), (3,4), (4,5) respectively, we can obtain that the equation F &; (x) = 0 has at least four real roots, but because F &; (x) is a polynomial of degree 4, there are at most four real roots, so the equation F &; (x) has four real roots; (x) = 0 has only four real roots and is located in the interval (1,2), (2,3), (3,4), (4,5) respectively

This is a polynomial function. Polynomial functions are continuous and differentiable on R. you have to prove it very quickly, but this is common sense. If you can prove that they are continuous and differentiable at any point, then according to the definition of interval continuous differentiable, they are continuous and differentiable on the whole interval. How can you feel unclear
All elementary functions: polynomials, exponents, logarithms, triangles and anti triangles are continuous and differentiable in their respective domain of definition, and their composite functions are generally continuous and differentiable, unless some meaningless points are defined as other values, which are artificially discontinuous or non differentiable, such as definition
If f (x) = sin (x) / X is 2 at the origin, the origin is discontinuous. However, if it is not the origin, because it is a composite function of elementary function, there is no problem of continuity and differentiability
To prove that it is differentiable in an interval, we only need to prove that every point in the interval is differentiable. If it is a closed interval, we can prove the existence of the right derivative for the left endpoint, and the existence of the left derivative for the right endpoint

If f (x) is differentiable and unbounded in the interval (a, b), then its derivative is unbounded in (a, b), but vice versa

1. Prove that the derivative function is unbounded. Let X and x0 belong to (a, b), and there exists y between X and x0, such that: F '(y) = f (x) - f (x0) / (x-x0). So | f' (y) | > = | f (x) | / (B-A) - | f (x0) / (B-A) |. Fix x0, because the function is unbounded, X can be changed so that | f (x) | is greater than any specified positive number, so | f '(y) |

How to judge whether a function is bounded or unbounded on an open interval
Sinin (1 + x) / (1-x) when the absolute value of X is greater than 1, is it bounded or unbounded?

The judgment of bounded and unbounded mainly depends on the point or boundary that tends to infinity or undefined