How to judge whether a function is differentiable and continuous in an interval in Calculus

Differentiable must be continuous, and continuous is not necessarily differentiable
Judge continuity: set point x0. If x tends to x0, limf (x) = f (x0), then f (x) is continuous at x0
Judgment can lead: left lead = right lead, defined by
LIM (f (x) - f (x0)) / (x-x0), where x tends to x0 + and x0-
For example, f (x) = |x|
To verify whether x = 0 is differentiable
When x tends to x0 +, LIM (f (x) - f (0)) / (x-0) = Lim X / x = 1
When x tends to x0 -, LIM (f (x) - f (0)) / (x-0) = LIM (- x) / x = - 1
So the left derivative is not equal to the right derivative, and the derivative of F (x) does not exist at point 0

Calculus function judgment 2、 Judgment question 1. The necessary and sufficient condition for the differentiability of univariate function is that the left and right derivatives exist and are equal A. Error B. correct 2. The cosine of the image of the function at a certain point is the geometric meaning of the derivative A. Error B. correct 3. Total differential of multivariate function u = XYZ + 2008 Du = 2008 + yzdx + xzdy + xydz A. Error B. correct 4. The original functions of power functions are power functions A. Error B. correct 5. The derivative expression of implicit function cannot contain y. () A. Error B. correct 6. The extreme point must be included in the internal stagnation point of the interval or the point where the derivative does not exist A. Error B. correct 7. The derivative or differential equation with unknown number is called differential equation A. Error B. correct 8. Univariate function must be continuous and continuous must be derivative A. Error B. correct 9. The derivative of the inverse function of a function is equal to the reciprocal of its derivative A. Error B. correct 10. Infinitesimal quantity is a very small quantity A. wrong B. correct

1B,2A,3A,4A,5A,6B,7A,8A,9A,10A.

University Mathematics calculus continuous function master Example in the book: prove that the x power of function y = e is a continuous function on (- ∞, + ∞). In the book, it is necessary to prove that the function is continuous on x = 0. Why? [example 4.4 on page p62 of calculus]

Yes;
In y = e ^ x, X is always negative when approaching from the left of 0;
X is always positive when approaching from the right of 0;
In many cases, the two limits are not the same. The negative approach method must change to the denominator, so it must be examined;
Power function, exponential function, this is to be considered. To learn this proof topic is to think like this in the future;

Calculus f e ^ 2x / 1 + e ^ 2x DX solution

∫e^(2x)/[1+e^(2x)] dx
= ½ ∫ 1/[1+e^(2x)] d(e^(2x))
= ½ ln(1+e^(2x))+C

Limx → 0 (e ^ x-cosx-2x) / x ^ 2-2x several AP calculus problems for advice! 1 find the value of this limit in the problem 2 if G (x) = ∫ (upper 2x lower 0) f (T) DT, find the value of G '(3)? The original question is The number of moose in a national park is modeled by the function m that satisfies the logistic differential equation DM / dt = 0.6m (1-m / 200), where t is the time in years and m (0) = 50, what is LIM (t → infinity) m (T)? 4 infinite series Σ (n = 1 to infinity) n / (n ^ P + 1) converges to find the value range of P 5 ∫2x/(x+2)(x+1) dx=? I think this problem should be solved by partial integral, but it is always bypassed = π

1. Rubida's law, after the upper and lower derivatives are obtained, limx → 0 (e ^ x + sinx-2) / (2x-2) fractions tend to - 1 and - 2 respectively, so the answer is 1 / 22. G '(x) = 2F (2x), G' (3) = 2F (6) 3. Solve the differential equation 200dm / [M (200-m)] = (0.6) DT [(1 / M) + 1 / (200-m)] DM = 0.6dtln [M / (200-m)] = 0

F (x) = (2x ^ 2-x) / (x-1) g (x) = 2x + 1 image property calculus F (x) = (2x ^ 2 – x) / (x-1) g (x) = 2x + 1 two images 1. F (x) is also equal to 2x + 1 + (1 / x-1); What does the image of this formula have to do with the original two formulas when x approaches + - infinity? 2. F (x) can also be written as (2x-1) / (1-1 / x); Is the image of this formula basically the same as H (x) = 2x-1 when x is close to + - infinity? Why? It should be explained by images

F '(x) = 2 + [1 / (x-1)]' when x tends to positive and negative ∞, [1 / (x-1)] 'tends to 0, so f' (x) tends to 2, and then f (x) - G (x) = 1 / (x-1) when x tends to positive and negative ∞, 1 / (x-1) tends to 0, that is, the two function images are infinitely close. In fact, G (x) is the asymptote of F (x), f (x) - H (x) = 2 + 1 / (x-1

How to judge whether a function is differentiable and continuous in an interval in Calculus