The derivative value of function y = f (x) at one point is 0, which is the extreme value of function y = f (x) at this point___ Conditions

The derivative value of function y = f (x) at one point is 0, which is the extreme value of function y = f (x) at this point___ Conditions

According to the definition of function extremum, when the differentiable function gets extremum at a certain point, f '(x) = 0 must hold. But when f' (x) = 0, the function does not necessarily get extremum, for example, function f (x) = X3. Function derivative f '(x) = 3x2, when x = 0, f' (x) = 0, but function f (x) = X3 monotonically increases, there is no pole

"The derivative value of the differentiable function y = f (x) at one point is 0" is the ()
A. Necessary and insufficient condition B. sufficient and unnecessary condition C. sufficient and necessary condition D. neither sufficient nor necessary condition

If the derivative value of the function y = f (x) at one point is 0, then the function y = f (x) does not necessarily take the extreme value at this point. For example, the function f (x) = X3 satisfies f '(0) = 0, but x = 0 is not the extreme value. If the function y = f (x) takes the extreme value at this point, then according to the definition of the extreme value, the derivative value of y = f (x) at one point is 0, which holds, and "the derivative value of the function y = f (x) at one point is 0" is "the function y = f (x) ）It is necessary and insufficient to take the extreme value at this point

Is it wrong that "the derivative value of function y = f (x) at one point is 0, which is the necessary condition but not the sufficient condition for function y = f (x) to take the extreme value at this point"?

How can I feel that it is neither sufficient nor necessary
First, the derivative of function f (x) at a point is zero, which does not mean that the extreme value can be taken at that point, for example, f (x) = x ^ 3
Second, the extreme value of a function at a point does not mean that the derivative of the function at that point is zero, provided that the function is at that point
Can be derived, and the derivative is 0, for example: y = sqrt (x ^ 2)

Solve the equation of the circle with the length of 4 √ 3, which is cut by two points P (4, - 2), q (- 1,3) tangent on the y-axis

The center perpendicular of PQ is x-y-1 = 0, so let the center o be (T, t-1), then the circular equation is (x-t) ^ 2 + (Y-T + 1) ^ 2 = │ op │ ^ 2 = (T + 1) ^ 2 + (T-4) ^ 2, let x = 0, and the solution is y = T-1 ± (T ^ 2-6t + 17) ^ (1 / 2). Because │ y1-y2 │ = 2 (T ^ 2-6t + 17) ^ (1 / 2) = 4 √ 3, T ^ 2-6t + 5 = 0, when t = 2 or 3T = 2, the circular equation is (...)

It is known that a circle passes through the coordinate origin and point P (1,1), and the center of the circle is on the line 2x + 3Y + 1 = 0

According to the intersection property of straight line and circle, the center of circle is on the perpendicular x + Y-1 = 0 of point O (0,0) and point P (1,1), and then according to the center of circle on the straight line 2x + 3Y + 1 = 0, the coordinates of center C are (4, - 3), so the radius r = | OC | = 5, so the equation of circle is (x-4) 2 + (y + 3) 2 = 25

It is known that a circle passes through the coordinate origin and point P (1,1), and the center of the circle is on the line 2x + 3Y + 1 = 0

According to the intersection property of straight line and circle, the center of circle is on the perpendicular x + Y-1 = 0 of point O (0,0) and point P (1,1), and then according to the center of circle on the straight line 2x + 3Y + 1 = 0, the coordinates of center C are (4, - 3), so the radius r = | OC | = 5, so the equation of circle is (x-4) 2 + (y + 3) 2 = 25

Find the equation of circle C whose center is on the line y = 2x and passes through the origin and point m (3,1)

If the center of the circle (a, 2a) is set, then | OC | = | om |, that is: A2 + (2a) 2 = (A-3) 2 + (2a-1) 2, the solution is a = 1, so the center coordinates (1, 2), radius is: 5. The equation of circle C: (x-1) 2 + (Y-2) 2 = 5

Solve the circle equation of the origin and point P (1,3) with the center of the circle on the straight line y = 2x + 4

Let the center of the circle be (x, 2x + 4)
（x-1)²+（2x+4-3)²=（x-0)²+（2x+4-0)²
If x = - 1, the center of the circle is (- 1,2)
Radius = √ [(- 1-0) & sup2; + (- 2 + 4-0) & sup2;] = √ 5
The circular equation is (x + 1) & sup2; + (Y-2) & sup2; = 5

Find the equation of circle C whose center is on the line y = 2x and passes through the origin and point m (3,1)

If the center of the circle (a, 2a) is set, then | OC | = | om |, that is: A2 + (2a) 2 = (A-3) 2 + (2a-1) 2, the solution is a = 1, so the center coordinates (1, 2), radius is: 5. The equation of circle C: (x-1) 2 + (Y-2) 2 = 5

Given the equation of circle: x ^ 2 + y ^ 2 = 1, find the tangent equation of point P (1,3) circle

I can calculate the distance between the two tangent lines (k-1) and the center of the circle (k-1)