The monotone decreasing interval of the function f (x) = the third power of X - the square of 3x + 2 is

The monotone decreasing interval of the function f (x) = the third power of X - the square of 3x + 2 is

f（x）=x^3-3x^2+2
f`(x)=3x^2-6x

Question 20: the function f (x) = the third power of 1 / 3x minus the square of 1 / 2 (a + 1 / a) x + X. first, if a is greater than 0, what is the value of a, the tangent slope of F (x) at point (1, f (1)) is the largest, and the tangent equation is obtained. Second, when a = 2, the function f (x) is not a monotonic function in the interval (K-3 / 4, K + 3 / 4), find the value range of the real number K. there is a process, thank you!

(1) When f '(x) = x ^ 2 - (a + 1 / a) x + 1, x = 1, f' (x) has a minimum value, indicating that (a + 1 / a) / 2 = 1, a = 1 (2) f (x) is not monotonic in the interval (K-3 / 4, K + 3 / 4), that is, f '(x) has both positive and negative values in this interval, indicating that f' (x) has a root in this interval. When a = 2, f '(x) = x ^ 2-5x / 2 + 1, f' (x) = 0

Let's know the function f (x) = the third power of ax - the square of 3x / 2 + 1 (x ∈ R), where the real number a > 0 (1). If a = 1, find Tangent equation of line y = f (x) at point (2, f (2)) (2) If f (x) > 0 is constant on the interval [negative half, half], find the value range of A

(1) When f (x) = ax ^ 3-3 / 2x ^ 2 + 1A = 1, f (x) = x ^ 3-3 / 2x ^ 2 + 1F '(x) = 3x ^ 2-3x tangent slope k = f' (2) = 12-6 = 6F (2) = 8-6 + 1 = 3, tangent point (2,3) tangent equation Y-3 = 6 (X-2), i.e. 6x-y-9 = 0 (2) x ∈ (- 1 / 2,1 / 2), f (x) > 0 is constant. When x = 0, f (0) = 1 > 0, a > 0 all conform to when 03 / (2x) - 1

Known function f (x) = 1 3x3 + AX2 + BX (a, B ∈ R) gets the extreme value when x = - 1 (1) Try to express B with an algebraic formula containing a; (2) Find the monotone interval of F (x)

(1) According to the meaning of the question, f '(x) = x2 + 2aX + B, because x = - 1 is an extreme point of the function,
Then f ′ (- 1) = 1-2a + B = 0, B = 2a-1;
(2) Because the function f (x) has extreme points, the equation f '(x) = 0 has two unequal real roots,
From (1), f '(x) = x2 + 2aX + B = x2 + 2aX + 2a-1 = (x + 1) (x + 2a-1),
Let f '(x) = 0, and the solution is X1 = - 1 or x2 = 1-2a,
① When X1 > X2, i.e. a > 1, the changes of F '(x) and f (x) are as follows:
Therefore, the monotonic increasing interval of function f (x) is (- ∞, 1-2a) and (- 1, + ∞), and the monotonic decreasing interval is (1-2a, - 1);
② When X1 < X2, i.e. a < 1,
Similarly, it can be obtained that the monotonic increasing interval of function f (x) is (- ∞, - 1) and (1-2a, + ∞), and the monotonic decreasing interval is (- 1, 1-2a)
To sum up, when a > 1, the monotonic increasing interval of function f (x) is (- ∞, 1-2a) and (- 1, + ∞), and the monotonic decreasing interval is (1-2a, - 1);
When a < 1, the monotone increasing interval of function f (x) is (- ∞, - 1) and (1-2a, + ∞), and the monotone decreasing interval is (- 1, 1-2a)

Let a belong to R, if the ax power of function y = e + 3x (x belongs to R), there is an extreme point greater than 0, find the range of A It's best to write down the complete solution process

From the meaning of the question, it is obtained that y = e ^ ax + 3x has a point with derivative 0 when x > 0;
y'=a*e^ax+3=0-》e^ax=-3/a; Because e ^ ax > 0, ax = ln (- 3 / a) / A; Because x > 0 exists; a

If f (x) is a periodic function with period T If f (x) is a periodic function with period T and f (x) has a center of symmetry (a, 0), then all points (a + kt / 2,0) k ∈ Z are the center of symmetry of F (x)

Because f (x) = f (x + KT)
There is a center of symmetry (a, 0), so f (x) + F (2a-x) = 0
So f (x + KT) + F (2a-x) = 0
And f (2a-x) = f [(2 (a + kt / 2) - (x + KT)]
So f (x + KT) + F [(2 (a + kt / 2) - (x + KT)] = 0
So the point (a + kt / 2,0) k ∈ Z is the center of symmetry of F (x)