The function f (x) = | x + 1 | + | ax + B | (B ≠ 1), if there are three unequal real numbers x1, X2, X3, such that f (x1) = f (x2) = f (x3), then a =?

The function f (x) = | x + 1 | + | ax + B | (B ≠ 1), if there are three unequal real numbers x1, X2, X3, such that f (x1) = f (x2) = f (x3), then a =?

Let Y1 = x + 1, y2 = ax + B
y1＝0 x1＝-1 y2＝0 x2＝﹙-b/a﹚
There are three unequal real numbers x1, X2, X3, such that f (x1) = f (x2) = f (x3) ←→ f (- 1) = f (- B / a)
It can be seen from the drawing that in many other cases, two point functions are equal
|a﹙-1﹚+b|＝|-b/a+1| a＝[b±√﹙b²±4b-4﹚]/2
According to the value of B, there are 0.1, 2, 3 and 4 a values

Let f (x) = x ^ 2-6x + 6, x > = 0,3x + 4, X

I can't understand your title. Is it wrong to write x > = 0? How can I still write x

Let f (x) = x ^ 4 + AX3 + 2x2 + B. if f (x) has an extreme value only at x = 0, find the value range of A

f'(x)=4x^3+3ax^2+4x=x(4x^2+3ax+4),
F "(x) = 12x ^ 2 + 6AX + 4, F" (0) = 4 > 0, so f (0) is a minimum
Only x = 0 is the extreme point, then the equation 4x ^ 2 + 3ax + 4 = 0 has no real root or equal root, that is delta = 9A ^ 2-4 * 4 * 4