Let f (x) = x ^ 3-6x + 5 If the equation f (x) = a about X has three different real roots, we can find the value range of real number a

1) Find the monotone interval and extremum of function f (x)
(2) If the equation f (x) = a about X has three different real roots, find the value range of real number a
(3) When x belongs to (1, positive infinity), f (x) > = K (x-1) is constant. The mathematical range of real number k is obtained
f(x)=x^3-6x+5
f'(x)=3x^2-6=0
x=±√2
When X - √ 2, f '(x) > 0, f (x) increases
-√2

Let f prime (x) = x3-4.5x2 + 6x-a
1. For any real number x, if f (x) is greater than or equal to m, the maximum value of M is obtained
2. If the equation f (x) = 0 has only one real root, the value range of a is obtained

Find the maximum value of M is the minimum value of the function (this problem must have a minimum value), so the derivative of the function = 0, the derivative function 3x ^ 2-9x + 6 = 0, that is, x ^ 2-3x + 2 = 0, get x = 1 and x = 2, when x = 1, the function = 1-4.5 + 6-A = 2.5-A, when x = 2, the function = 8-18 + 12-A = 2-A, so when x = 2, it is the minimum value of the function, the maximum value of M is the maximum value of the function

Given function f (x) = x3-3x2 + 1 through point P (0,1) tangent equation
The function f (x) = x3-3x2 + 1 is known
Finding the tangent equation of point P (0,1)

f(x)=x3-3x2+1
The derivative is f '(x) = 3x ^ 2-6x
Substitute x = 0 to get k = 0
So the tangent equation is y = 1

The function f (x) = x-3x2 + 1 has a minimum at x =
Isn't it impossible to get a minimum?

-If it's 3x squared, there's no minimum, there's only a maximum
And there's no range of X, and if there is, there may be a minimum

Let f (x) = x ^ 3-6x + 5 If the equation f (x) = a about X has three different real roots, we can find the value range of real number a