It is proved that the function f (x) 2 (3x-l) / X is infinite when x → o

It is proved that the function f (x) 2 (3x-l) / X is infinite when x → o

3x_ 1 is the same medium infinity of X

It is proved that the function f (x) = 3x ^ 4 is an increasing function in the interval [0, + ∞)

The definition of senior one proves that:
Let X1 > x2 > = 0
f(x1)-f(x2)=3(x1^4-x2^4)=3(x1^2+x2^2)(x1+x2)(x1-x2)
Because x1-x2 > 0, X1 + x2 > 0
So f (x1) - f (x2) > 0
That is, f (x1) > F (x2)
Therefore, the function f (x) = 3x ^ 4 is an increasing function in the interval [0, + ∞)

Derivative function of F (x) = (3x ^ 2-3x-3) e ^ - x

f'(x)=(6x-3)*e^(-x)+(3x²-3x-3)*e^(-x)*(-x)'
=(6x-3)*e^(-x)-(3x²-3x-3)*e^(-x)
=(-3x²+x)*e^(-x)

Finding the derivative function of F (x) = 3x ^ 2 + X

f'(x)=[3x^2+x]'
=6x+1

Known function f (x) = 1 / 3x ^ 3-ax ^ 2 + (a ^ 2 + 1) x + B (a, B belong to real number)
When a is not equal to 0, if f (x) is not monotone in the interval (- 1,1), the value of a is obtained

This problem should only work out the range of a, not wait until the specific value
Because f (x) is not monotone on the interval (- 1,1)
Then the derivative of F (x) has zeros in the interval (- 1,1)
f'(x)= -1/4x -2ax +a ^2+1
f'(1)=a^2 -2a f'(-1)=a^2 +2a
And a is not equal to 0, f '(1) f' (- 1) = a ^ 2 (a + 2) (A-2)

If α∈ R, f (x) = ax ^ 3-3x ^ 2 (1). If a = 1, find the extremum and monotone interval of F (x)

When a = 1,
f(x)=x^3-3x^2
f '(x)=3x^2-6x=3x(x-2)
Let f '(x) = 0, then X1 = 0; x2 = 2 are two stable points of F (x);
When X0
So, X2 = 2 is the minimum point of the function;
F (min) = f (2) = 8-12 = - 4;
Let f '(x) > 0 = = > x > 2; or X

Let f (x) = x ^ 3-ax ^ 2-3x. If x = 3 is the extreme point of F (x), find the monotone interval of F (x)

F "(x) = 3x ^ 2-2ax-3, let x = 3 get 0 = 24-6a, a = 4, Let f" (x) = 0 get x = 3 or - 1 / 3

If the function y = - 4 / 3x ^ 2 + ax has three monotone intervals, then the value range of a is, to process. Thank you

Wrong
It is y = - 4 / 3x & sup3; + ax
y'=-4x²+a
There are three monotone intervals
That is, increase or decrease, increase or decrease
So the sign of Y 'is + - + or-+-
So there are two intersections of the Y 'and X axes
So the discriminant = 0 + 16A > 0
a>0

Let f (x) = - 1 / 3x ^ 3 + ax have three monotone intervals, then the value range of a is

According to the meaning of F (x) [at least] there is an extreme point in the interval (- 1,1),
Since f '(x) = 3x ^ 2 2 (1-A) x-a (a 2) = (x-a) [3x (a 2)],
When a ≠ - 1 / 2, f (x) has two different extreme points X1 = A and X2 = - (a 2) / 3,
① When a = - 1 / 2, f (x) increases strictly monotonically
②-1

Given that the function y = 1 / 3x ^ 3 + x ^ 2 + AX-5 is a monotone increasing function on [1, positive infinity], then the value range of a is

Given that the function y = (1 / 3) x & # 179; + X & # 178; + AX-5 is a monotone increasing function on [1, + ∞), then the value range of a is
If y = (1 / 3) x & # 179; + X & # 178; + AX-5 is a monotone increasing function on [1, + ∞), its derivative y '= x & # 178; + 2x + a = (x + 1) &# 178; - 1 + a must be satisfied
Y '(1) = 3 + a ≥ 0, i.e. a ≥ - 3. When a ≥ - 3, the function y is monotone increasing on [1, + ∞)