F (x) = sgnx in high numbers. What function is sgnx here? Or what? How to use it? graphical

F (x) = sgnx in high numbers. What function is sgnx here? Or what? How to use it? graphical

Piecewise function
Greater than zero is 1, less than - 1, equal to 0
Mainly useful in circuits

Define "symbolic function" f (x) = sgnx = {① 1, x > 0 ② 0, x = 0 ③ - 1, x < 0, then the solution set of inequality: x + 2 > (X-2) ^ sgnx is?

x> 0, x + 2 > X-2, 2 > - 2 is constant, so x > 0
X = 0, x + 2 > 1, x > - 1, so x = 0
X < 0, x + 2 > 1 / (X-2), x ^ 2-4 < 1, - root 5 to sum up: x > - root 5

Derivative f '(x) approximate shape of image How to draw the derivative of a function? For example, what is the derivative of y = x + 1?

Flower image after derivation
The derivative of y = x + 1 is y = 1, and the image is a straight line parallel to the X axis

Let the symbolic function f (x) = sgnx = {① 1, x > 0, ② 0, x = 0, ③ - 1, X

X + 1 = {① X-2, x > 0 ② 1, x = 0 ③ 1 / (X-2), x0 x + 1 = X-2 = > 1 = - 2 = > ∅ ② x = 0 x + 1 = 1 = > x = 0 ③ x + 1 = 1 / (X-2) = > x ^ 2-2x + X-2 = 1 = > x ^ 2-x-3 = 0 = > (x-1 / 2) ^ 2-1 / 4-3 = 0 = > x = + - sqrt (13 / 4) + 1 / 2, the solution set is {0, - sqrt (13 / 4) + 1 / 2, sqrt (13 / 4) + 1 /

It is proved that the function y = 1 / xsin1 / X is unbounded in (0,1)

Should the title be f (x) = (1 / x) * sin (1 / x)? I did it
According to Heine theorem, as long as we find an unbounded sequence, we can prove that the function is unbounded
So take the sequence: xn = 1 / (2n π + π / 2),
Then xn ∈ (0,1), and the limit of XN is 0, and the limit of corresponding f (xn) is positive infinity
That is, the function is unbounded in this interval

It is proved that the function f (x) = 1 / xsin1 / X is unbounded in the interval (0,1], but when x approaches 0 + 0, the function is not infinite

1) Prove unbounded. Let xn = 2 / (4N + 1) π, then f (xn) = (4N + 1) π / 2 * sin (4N + 1) π / 2 = (4N + 1) π / 2. When xn tends to 0, f tends to infinity, so it is unbounded. We can use the well-defined proof 2) and let xn = 1 / 2n π, then f (xn) = 2n π * sin2n π = 0. According to 1), when x tends to 0, f is