The function f (x) = AX-1 - ㏑ x.1 is known. The number of extremum points of the function in the domain of definition is discussed. 2. F (x) obtains extremum at x = 1, and for any x ∈ (0, ∞)

The function f (x) = AX-1 - ㏑ x.1 is known. The number of extremum points of the function in the domain of definition is discussed. 2. F (x) obtains extremum at x = 1, and for any x ∈ (0, ∞)

f（x）=ax-1-㏑x
When a = 0, there is no extremum. When a is not equal to 0, the extremum is 1 / A
If f (x) has an extreme value at x = 1, then a = 1

Given that the function FX = ax ∧ 3 BX ∧ 2 CX has the extremum when x = ± 1, and f (1) = - 1, the analytic expression of the function FX is obtained
Finding monotone interval of function FX

If FX = ax ∧ 3 + BX ∧ 2 + CXF 'x = 3ax ∧ 2 + 2bx + CX = ± 1, we get the extremum, so x = ± 1 is the root of 3ax ∧ 2 + 2bx + C = 0, so 0 = - 2b / (3a) - 1 = C / (3a) and f (1) = - 1, we get - 1 = a + B + C, we get a = 1 / 2B = 0C = - 3 / 2, the analytic expression of function FX is FX = 1 / 2x ∧ 3 - 3 / 2cx (2) f' x = 3 / 2x ∧ 2-3 / 2

It is known that the function f (x) = X3 + bx2 + CX + 2 has extremum when x = 1. (I) find the value of B and C; (II) if the function f (x) = X3 + bx2 + CX + 2 has extremum when x = 1（
It is known that the function f (x) = X3 + bx2 + CX + 2 has an extreme value of 6 when x = 1
(I) finding the value of B and C;
(II) if the tangent of F (x) is parallel to the line 3x + y + 1 = 0, the tangent equation is obtained
We can see that f ′ (x) = 3x2-12x + 9,
The slope of the tangent is - 3. How can the slope be - 3

The line 3x + y + 1 = 0 can be changed into y = - 3x-1, the slope is - 3, and the slopes of parallel lines are equal

Given the function f (x) = X3 + bx2 + CX + 2, we can get the extreme value - 1 at x = 1. (1) find the value of B and C; (2) if the equation f (x) + T = 0 about X has a real root in the interval [- 1,1], find the value range of real number t

(1) Let g (x) = f (x) + T = X3 + x2-5x + 2 + T, then G ′ (x) = 3x2 + 2x-5 = (3x + 5) (x-1) (1) (1) (1) (2) (1) (2) (1) (1) (1) (2) (1) (1) (1) (2) (1) (1) (2) (1) (1) (1) (2) (1) (1) (1) (1) (1) (1) (1) (1) (1) (1) (1) (1) (2) (1) (1) (1) (1) (1) (1) (2) (1) (1) (1) (2) (1) (1) (1) (1) (2) (1) (1) (2) (1) (1) (1) (2) (1) (1) (1) (1) (1) (2) (1) (1) (1) (1) (1) (1) (1) (1) (1) (1) (1

Given the function f (x) = - 13x3 + bx2 + CX + BC about X, if the function f (x) has the extreme value - 43 at x = 1, try to determine the value of B and C

F '(x) = - x2 + 2bx + C, ∵ f (x) has extremum at x = 1 - 43 ∵ f (1) = - 13 + B + C + BC = - 43, f' (1) = - 1 + 2B + C = 0, the solution is: B = 1, C = - 1 or B = - 1, C = 3

The known function f (x) = 1 / 3x ^ 3-1 / 2 (2a + 1) x ^ 2 + (a ^ 2 + a) x
The known function f (x) = 1 / 3x ^ 3-1 / 2 (2a + 1) x ^ 2 + (a ^ 2 + a) x
(1) If f (x) has a maximum at x = 1, find the value of real number a;
(2) If any m ∈ R, the line y = KX + m is not the tangent of the curve y = f (x), the value range of K is obtained;
(3) If a ＞ - 1, find the maximum value of F (x) in the interval [0,1].

(1) If f '(x) = x ^ 2 - (2a + 1) x + A ^ 2 + A, Let f' (x) = 0, then x = a or x = a + 1. From the meaning of the question, we know that f (x) increases monotonically on (- ∞, a), (a + 1, + ∞) and decreases monotonically on (a, a + 1), so f (x) has a maximum at x = a, so a = 1
(2) From the meaning of the question, f '(x) = k has no solution, that is, x ^ 2 - (2a + 1) x + A ^ 2 + A-K = 0 has no solution, so (2a + 1) ^ 2-4 (a ^ 2 + A-K) = 4K + 1 < 0, K < - 1 / 4
(3) When - 1 ＜ a ＜ 0, f (x) decreases monotonically on [0, a + 1], increases monotonically on (a + 1,1), f (0) = 0, f (1) = a ^ 2-1 / 6, so when - 1 ＜ a ≤ - √ 6 / 6, f (x) max = f (1) = a ^ 2-1 / 6, when - √ 6 / 6 ＜ a ＜ 0, f (x) max = f (0) = 0
When 0 ≤ a < 1, f (x) increases monotonically on [0, a] and decreases monotonically on (a, 1), so f (x) max = f (a) = 1 / 3A ^ 3 + 1 / 2A ^ 2
When a ≥ 1, f (x) increases monotonically in the interval [0,1], so f (x) max = f (1) = a ^ 2-1 / 6

Given f (x + 1) = x2-3x + 2 (1), find the values of F (2) and f (a); (2) find the analytic expressions of F (x) and f (x-1)

(1) Let x + 1 = t, then x = T-1, f (T) = (t-1) 2-3 (t-1) + 2 = t2-5t + 6. The analytic formula of the function is f (x) = x2-5x + 6, (1) f (2) = 22-5 × 2 + 6 = 0, f (a) = a2-5a + 6, (2) from (1), f (x) = x2-5x + 6, f (x-1) = (x-1) 2-5 (x-1) + 6 = x2-7x + 12

Given the function f (x) = | x ^ 2-2 |, if f (a) = f (b), and 0 < a < B, then the minimum distance from the moving point P (a, b) to the straight line 3x + 4y-15 = 0 is

First of all, the value range of X and y can be determined as [- 2,2] according to the image. Therefore, by studying the distance formula | 3x + 4y-15 | / 5, we can know that the value of 3x + 4Y can not be greater than 15. Therefore, to make the distance shortest, that is to find the maximum value of 3x + 4Y. Therefore, there must be x in [0, root 2], y in (root 2,2]. Then write the expressions of the function in these two intervals respectively. According to f (a) = f (b), an equation can be obtained, After simplification, it is a * 2 + b * 2 = 4, which is the distance from the point on the circle to the straight line. But pay attention to the value range of a and B. if it's not easy to express, don't write it. You can calculate it yourself

Y = x + A ^ 2 / X (a is greater than 0) for function extremum

Because f '(x) = 1-A ^ 2 / x ^ 2, if f' (x) = 0, the solution is: x = - A or x = A and a > 0f '(x) > 0, the solution is: Xa
Let f '(x)

The extremum of function y = e ^ x ^ 2 is X=

y = e^(x^2)
y' = 2x .e^(x^2) =0
x=0
y'' = 2e^(x^2) [ 1+ 2x^2]
y''(0) > 0 (min)
min y = y(0) =1
The extreme point is x = 0