If the derivative of F (x) is continuous at x = A and limf '(x) / x-a = - 1 (when X -- > A), then A x = a is the minimum point B x = a is the maximum point C f '' (a) does not exist D f '' (a exists and is 0

If the derivative of F (x) is continuous at x = A and limf '(x) / x-a = - 1 (when X -- > A), then A x = a is the minimum point B x = a is the maximum point C f '' (a) does not exist D f '' (a exists and is 0

B
Limf '(x) = 0, X - > A; and the derivative of F (x) is continuous at x = a, so f' (a) = 0;
And f '' (a) = LIM (f '(x) - f' (a)) / (x-a) = limf '(x) / (x-a) = - 1

Derivative of F (x) = (x-a) ^ 1 / 2

f'(x)= 1/2*(x-a)^(-1/2)

There are four different intersections between F (x) = 2x ^ 2 + m and G (x) = ln | x | to find the value range of M~~

Because f (x) and G (x) are even functions, there are two different intersections when x > 0
The tangent slopes of F and G are 4x and 1 / x, respectively,
When the two curves are tangent, 4x = 1 / x, x = 1 / 2, that is, M = - ln2-1 / 2 when they are tangent,
Can get m

Thank you for helping me

It should be halves

Plural of half

halves
The plural of a noun ending in F is to change f to V and add es
And knif -- knives

Deepen the understanding of the value of non negative real number x "rounded" to the individual bit, and record it as 〈 x 〉
When n is a nonnegative integer, if n-1 / 2 ≤ x ∠ n + 1 / 2,
Then 〈 x 〉 = n
For example, < 0 > = < 0.48 > = 0, < 0.64 > = < 1.493 > = 1, < 2 > = 2, < 3.5 > = < 4.12 > = 4
Try to solve the following problems:
(1) fill in the blanks: 〈 x 〉=__ (π is the circumference ratio);
(2) if 〈 2x - 1 〉 = 3, find the value range of real number X

7/4<=X>9/4

When n is a non negative integer, if n-12 ≤ x ≤ n + 12, < x > = n. for example, < 0 > = < 0.48 > = 0, < 0.64 > = < 1.493 > = 1, < 2 > = 2, < 3.5 > = < 4.12 > = 4 If ＜ X-1 ＞ = 3, then the value range of real number x is___ ．

According to the meaning of the question, X-1 ≥ 2.5x-1 < 3.5, the solution is 72 ≤ x < 92

How to calculate the value of non negative real number x "rounded" to one digit
The value of non negative real number x "rounded" to one bit is recorded as: when n is a non negative integer, if n-1 / 2

2 0 1

How does matlab carry on Hilbert transform to a signal?
By the way, tell me where the program is wrong
t=0:pi/50:2*pi;
x=sin(t);
size=(x);
y=hilbert(x);
plot(t,y,'r');
Originally the result should be a cosine function, but the result is a sine

Warning:Imaginary parts of complex X and/or Y arguments ignored
The x-axis of the image is the time t, and the y-axis. Matlab removes the imaginary part of the Hilbert transform by default. You know that the real part of the Hilbert transform does not change, so there is no difference between the drawn graph and sin (T)

What is the Hilbert transformation of COS (w0t)

When w is real and t is imaginary: cos (WT) (LN (- t) - ln (T)) / PI + sin (ABS (W) * t)