Given f (x) = e ^ x-1-x-ax ^ 2, when x ≥ 0, f (x) ≥ 0, find the range of A

Given f (x) = e ^ x-1-x-ax ^ 2, when x ≥ 0, f (x) ≥ 0, find the range of A

F (x) = e ^ x-1-x-ax ^ 2, f '(x) = e ^ x-1-2ax when a ≤ 1 / 2, f' (x) = e ^ X-1 ≥ 0 is constant when x ≥ 0, f (x) increases monotonically in [0, + ∞), and f (0) = 0. So when a ≤ 1 / 2, f (x) ≥ 0 when x ≥ 0, when a > 1 / 2, f (x) = e ^ x-1-x-ax ^ 2 is decomposed into H (x) = e ^ X-1 and G (x) = x + ax ^ 2, that is, f (x) =

The maximum value and derivative of mathematical function in Senior Two
Let f (x) = ax ^ 3-6ax ^ 2 + B have the maximum value of 3 and the minimum value of - 29 in the interval [- 1,2], and a > 0
The answer is a = 2, B = 3

f(x)=ax^3-6ax^2+b
f'(x)=3ax^2-12ax
=3ax(x-4)
Because a > 0
So f (x) increases in the interval (negative infinity, 0) and decreases in the interval (0,4)
f(-1)=-7a+b,f(2)=-16a+b
So the minimum value in [- 1,2] is f (2) = - 16A + B = - 29
Maximum f (0) = b = 3
The solution is a = 2, B = 3

The monotonicity of the function in the first year of senior high school
f(x)=x+a/x(a>0)

Let X1 > x2 > 0, so f (x1) - f (x2) = x1-x2 + A / x1-a / x2 = x1-x2 + a (x2-x1) / X1 * x2 = (x1-x2) (x1 * x2-a) / X1 * x2
Therefore, when X1 * x2 > A, the substitution increases, and when X1 * x2 > x2 * x2 > root a, the substitution holds, so when x is greater than or equal to root a, the substitution increases, when x is greater than or equal to root a, and when x is less than or equal to root a, the substitution decreases,
For X