The function y f x is continuous at the point x0 and it is derivable at x0 .

Option C, requisite. If continuous but not necessarily derivable.

Conductor must be continuous.

Proof that the function f(x) is derivable at x0 and f(x) is defined in the x0 domain.

For an arbitrarily small >0, there is x=1 [2f'(x0)]>0 that makes :

[f(x0+⊿x)-f(x0)<ε

This can be deduced from the derivative definition.

A modern definition of a function.

is given a set of numbers a, assuming that the element in it is x, and the corresponding law is applied to the element x in a.

f, denoted as f(x), to get another set of numbers b, assuming that the element in b is y, then the equivalence between y and x can be expressed by y=f(x), and the concept of function has three elements: defining the domain.

a. Value range. b and the corresponding law f. The core of this is the correspondence law f, which is the essential feature of functional relations.

The function y f(x) is continuous at the point x0 and is necessary for it to be derivable at x0.

If the domain of a function is all real numbers, that is, the function is defined on the field of real numbers, then a certain condition is required for the function to be derivable at one point in the defined domain. First, to make the function f derivable at a point, the function must be continuous at that point. In other words, if a function is derivable at a point, it must be continuous at that point.

Derivable functions must be continuous, and discontinuous functions must not be derivable.

The function y=f(x) is continuous at the point x0 is a necessary condition for it to be derivable at x0, which is necessarily continuous, and continuous is not necessarily derivable.

The function is continuous at that point and both the left and right derivatives exist and are equal. Functions are derivative, and functions are continuous; Function continuity is not necessarily derivable; Discontinuous functions must not be derivative.

Not all functions have derivatives, and a function does not necessarily have derivatives at all points. If a function exists at a certain point in derivative, it is said to be derivable at that point, otherwise it is called underivable. However, the derivable function must be continuous; Discontinuous functions must not be derivative.

From the relationship between derivable and continuous: "derivable must be continuous, continuous is not necessarily derivable", it can be seen that the function f(x) is continuous at the point x=x is a necessary but not sufficient condition for f(x) to be derivable at x.

The sufficient and necessary conditions for a function to be derivable at a point are that the left and right derivatives are equal and continuous at that point. Obviously, if the function has "vertices" within the interval, (e.g. f(x)=|x|x=0 point) then the function is not derivative at that point.

c Conductible must be continuous, continuous is not necessarily directable.

Summary. Hello, continuity may not be leading.

Continuity is a necessary condition for derivability;

Conductivity is a sufficient condition for continuity.

If the function y=f(x) is continuous at x=x0, is the function y=f(x) derivable at x=x0?

Hello, continuity may not be leading. Continuity is a necessary condition for derivability; Conductivity is a sufficient condition for continuity.

Supplementary information: From "the function y=f(x) is continuous at x=x0", the bending rubber cannot be deduced "the function y=f (laughing height x) is derivable at x=x0", e.g. the function y=|x|Continuous at x=0, but not derivable, and from "the function y=f(x) is derivable at x=x0", it can be obtained that "the function y=f(x) is continuous at x=x0", so "the function y=f(x) is continuous at x=x0" is a necessary buried inadequate condition for "the function y=f(x) is derivable at x=x0".

Analysis: To prove that f(x) is continuous at point x 0, it must be proved that <>f(x)=f(x 0).According to the definition that the function is derivable at the point x 0, two transformations are implemented step by step:

One is the transformation of trends; The second is the transformation of the form (which becomes the derivative-defined).

Proofing 1: Let x=x0 +δx then when x x 0, δx 0

f(x 0 +δx)

［f(x 0 +δx)-f(x 0 )+f(x 0 )］

·δx+f(x 0 )］

<>x+ <

f(x 0 )

f′(x 0 )·0+f(x 0 )=f(x 0 ).

The function f(x) is continuous at the point x 0.

Argument 2: The function f(x) is derivable at the point x 0 and has at the point x 0.

［f(x)-f(x 0 )］

y= <>

·δx)<>

<>x=f′(x 0 )·0=0.

f(x)=f(x 0 ).

The function f(x) is continuous at the point x 0.

Answer]: B Analysis] From the relationship between derivable and continuous: "Derivable must be continuous, continuous is not necessarily derivable", it can be known that B should be chosen

Answer]: Proof may wish to set the empty sum f(x0)>0Since f(x) is continuous at x0, there is a decentered neighborhood of x0 by the local number preservation theorem of the shed judgment limit, such that when x f(x) > 0 and when xu(x0), f(x) > 0

That is, there is a neighborhood of x0 u(x0), and when the bucket is scattered x(x0), f(x)0

Let f(x)=1 (ax+b)=(ax+b) (1).

f'(x)=-a(ax+b)^(2)

f''(x)=(1*2)a^2*(ax+b)^(3)f'''x)=-1*2*3)a^3*(ax+b)^(4)f^(n)(x)=(a)^n*n!*(ax+b)^(n-1)Derivatives

If the function y=f(x) is derivable at every point in the open interval, then the function f(x) is said to be derivable in the interval. At this time, the function y=f(x) corresponds to a definite derivative value for each definite x value in the interval, which constitutes a new function, which is called the derivative of the original function y=f(x), denoted as y'、f'(x), dy dx, or df(x) dx, referred to as the derivative.

Derivatives are an important pillar of calculus. Newton and Leibniz contributed to this feast. The function y=f(x) is the derivative f at the point x0'Geometric meaning of (x0):

Represents the slope of the tangent of the function curve at the point p0(x0,f(x0)) (the geometric meaning of the derivative is the tangent slope of the function curve at this point).