Is the breakpoint of a removable function differentiable? The left and right limits and the left and right derivatives of the removable function exist and are equal at the discontinuity It is said in the book about the unilateral derivative that f (x) is differentiable at x0 if and only if the left and right derivatives of F (x) at x0 exist and are equal. Then the separable function should be differentiable at the discontinuous point It is said in the book that f (x) is differentiable at x0, and f (x) must be continuous at x0. Then the function that can be removed should not be differentiable at the discontinuous point When I do a question, I will certainly think that discontinuity is not derivable. But these two statements are contradictory,

Is the breakpoint of a removable function differentiable? The left and right limits and the left and right derivatives of the removable function exist and are equal at the discontinuity It is said in the book about the unilateral derivative that f (x) is differentiable at x0 if and only if the left and right derivatives of F (x) at x0 exist and are equal. Then the separable function should be differentiable at the discontinuous point It is said in the book that f (x) is differentiable at x0, and f (x) must be continuous at x0. Then the function that can be removed should not be differentiable at the discontinuous point When I do a question, I will certainly think that discontinuity is not derivable. But these two statements are contradictory,

The definition of left and right derivatives is: Lim [f (x) - f (x0)] / (x-x0) X -- > x0 + or - if you check this definition, you will immediately find that there are no left and right derivatives that can go to the discontinuity point. I know what you mean by f '(x0 +), f' (x0 -), these two are not left and right derivatives, they are left and right derivatives of derivative function at x0

Why can the discontinuity of derivative function only be the second kind of discontinuity?

If f '(x0) exists, then f' (x0) = Lim [f (x) - f (x0)] / (x-x0) exists (both left and right approaches exist and are equal). If f '(x) is a jump breakpoint at x = x0, then Lim left approach f' (x) is not equal to Lim right approach f '(x), and lim left approach [f (x) - f (x0)] / (x-x0) = Lim right approach [f (x) -

Why the discontinuity of derivative function must be the second kind of discontinuity

Walk on the windowsill Level 2 May 5, 2011 query on the conclusion that the discontinuity point of derivative function must be the second kind of discontinuity point browse times: 290 reward points: 0 | solution time: 2010-12-5 18:01 | cyd1990 | prosecution since its derivative function has the second kind of discontinuity point, it means the left derivative of the point

Variance calculation
8 6 9 5 10 7 4 7 9 5 \7 6 5 8 6 9 6 8 8 7

1.8 6 9 5 10 7 4 7 9 5
The average is 7
The variance calculated by the formula is 3.6
2.7 6 5 8 6 9 6 8 8 7
Average = 7
The variance calculated by the formula is 1.4

Variance operation
If the average of sample x1, X2 and X3 is 8 and the variance is 2, how to find the average and variance of sample 3x1 + 1,3x2 + 1,3x + 3?

If the mean of sample x1, X2, X3 is 8 and the variance is 2, then the mean of sample 3x1 + 1,3x2 + 1,3x + 3 is 3 * 8 + 1 = 25 and the variance is 2 * 3 & sup2; = 18
If the same number is added to a group of data and the variance is not changed, and it is multiplied by a, then the variance becomes a & sup2; times of the original

Help me calculate the variance and median of this set of data
No calculator
A 9 5 7 8 7 6 7 7 7
B 24 6 8 7 8 9 10

A mean of 7, variance of 1.2, median of 7
The mean of B was 7, the variance was 5.4, and the median was 7.5

How to calculate variance? What is variance? I don't know much about it. It's better to have examples and concepts of analysis and variance

In probability theory and mathematical statistics, variance is used to measure the degree of deviation between a random variable and its mathematical expectation. In many practical problems, it is of great significance to study the degree of deviation between a random variable and its mean value

The vertex coordinates of quadratic function y = (K + 1) x & # 178; + K & # 178; - K are (0,2), and the value of K is obtained

The vertex coordinates of quadratic function y = (K + 1) x & # 178; + K & # 178; - K are (0,2)
Then K & # 178; - k = 2, K + 1 ≠ 0
So (K + 1) (K-2) = 0, K ≠ - 1
So k = 2
If you don't understand, I wish you a happy study!

How much is 13589236 divided by 345.9

39286.603

Without a computer, compare the size of √ 7 + √ 8 and √ 2 ×√ 6, and explain the reason