If the function f (x) = x · (x-C) 2 has a maximum at x = 2, then the value of constant C is () A. 6B. 2C. 2 or 6D. 23

If the function f (x) = x · (x-C) 2 has a maximum at x = 2, then the value of constant C is () A. 6B. 2C. 2 or 6D. 23

∵ f ′ (x) = (x-C) 2 + 2x (x-C), ∵ function f (x) = x · (x-C) 2 has a maximum at x = 2, ∵ (2-C) 2 + 4 (2-C) = 0, the solution is C = 2 or C = 6; after testing, C = 6, so a
Let f (x) = - 1 / 3x3 + 2ax2-3a2x + 1,0
f(x)=(-1/3)x³+2ax²-3a²x+1
The domain of the function is R. obviously, the function is continuous and differentiable in the domain
f'(x)=-x²+4ax-3a²
Let f '(x) = 0, then:
-x²+4ax-3a²=0
(x-a)(x-3a)=0
Therefore:
X = a or 3a
1）
When a
(1) : y '= - X & # 178; + 4ax-3a & # 178;, let y' = 0
→x1=a,x2=3a
When A0, f (x) ↑
So: x = a, f (a) is the maximum
f(a)=1-4a³/3
(2):-a≤f`(x)≤a
→f`(x)≤|a|
→f`(x)≤ax,f`(x)>0,f(x)↓
So: x = a, f (a) is the minimum
f(a)=1-4a³/3
Now 0
Given the function f (x) = (2 / 3) x ^ 3-2ax ^ 2 + 3x (x ∈ R), find the monotone interval of function f (x)
This problem is a typical application of derivative function monotone interval problem
It is known that f '(x) = 2x ^ 2-4ax + 3 = 2 (x-a) ^ 2-2a ^ 2 + 3
When - 2A ^ 2 + 3 ≥ 0, that is - √ 6 / 2 ≤ a ≤ √ 6 / 2, f '(x) ≥ 0, the function is an increasing function on R, that is, its monotonic increasing interval is r, and there is no decreasing interval; when - 2A ^ 2 + 3
Check carefully again to see if there is a wrong symbol or less condition.
I have a question here for reference only.
Given that the function f (x) = (2 / 3) x ^ 3-2ax ^ 2-3x decreases monotonically in the interval [- 1,1], find the value range of real number a
The derivative of the original function f '= 2x ^ 2-4ax-3
Monotonically decreasing on interval [- 1,1], then f '
Given that the function f (x) = x ^ 3 + ax ^ 2 - (A-1) + 7 has a maximum and a minimum, find the value range of A
The function f (x) = x ^ 3 + ax ^ 2 - (A-1) x + 7 has a maximum and a minimum, indicating that f '(x) = 0 has two solutions
F '= 3 * x ^ 2 + A * 2 * x - (A-1), there are two solutions, so the discriminant is > 0
Discriminant = (a * 2) ^ 2-4 * 3 * [- (A-1)]
=4*a^2+12*a-12>0
Solve the inequality by yourself
The function f (x) = - x ^ 2 + ax + 1-lnx is known. If the function f (x) has both maximum and minimum, the value range of a is obtained;
f ‘(x)=-2x+a-1/x=（ -2x²+ax-1）/x
Let f '(x) = 0, that is - 2x & # 178; + AX-1 = 0
If f (x) has both maxima and minima, then the equation - 2x & # 178; + AX-1 = 0, △ = A & # 178; - 8 > 0 leads to a ＞ 2 √ 2 or a ＜ - 2 √ 2
Ask again if you don't understand
f(x)=-x^2+ax+1-lnx
f'(x)=-2x+a-1/x=0
∵x≠0
∴-2x^2+ax-1=0
In order to make the function have extremum, the function must have inflection point, that is, the equation has two unequal real roots.
△=a^2-8≥0
A ≥ 2 root sign 2
The function f (x) = KX + 1x2 + C (c > 0 and C ≠ 1, K ∈ R) has a maximum point and a minimum point, one of which is x = - C. (1) find another extreme point of function f (x); (2) find the maximum value m and minimum value m of function f (x), and find the value range of K when M-M ≥ 1
(1) F ′ (x) = K (x2 + C) - 2x (KX + 1) (x2 + C) 2 = - kx2-2x + CK (x2 + C) 2, f ′ (- C) = 0, that is, c2k-2c-ck = 0, (*) ∧ C ≠ 0, ∧ K ≠ 0. From F ′ (x) = 0, - kx2-2x + CK = 0, another extreme point is x = 1 (or x = c-2k) from Weida's theorem. (II) from (*) formula, k = 2c-1, that is, C = 1 + 2K. When C > 1, K > 0; when 0 < C < 1, K < - 2. (I) when k > 0 The solution is k ≥ 2 from M-M = K2 + K22 (K + 2) ≥ 1 and K ＞ 0. (II) when k ＜ - 2, f (x) is an increasing function in (- ∞, - C) and (1, + ∞), and a decreasing function in (- C, 1) 22 (K + 2) > 0, M = f (1) = K2 < 0, M-M = - K22 (K + 2) - K2 = 1 - (K + 1) 2 + 1K + 2 ≥ 1. In conclusion, the range of K is (- ∞, - 2) ∪ [2, + ∞)
Let f (x) = ax2ex, where a ≠ 0. (I) find the derivative of F (x); (II) find the maximum of F (x)
(1) F ′ (x) = axex (x + 2), (II) we know from (I): F ′ (x) = axex (x + 2), (I) when a > 0, when f ′ (x) > 0, we get x > 0 or x < - 2; when f ′ (x) < 0, we get - 2 < x < 0; the monotone decreasing interval of F (x) is (- 2,0); the monotone increasing interval of F (x) is (- ∞, - 2) and (0, + ∞). (5 points), so when x = - 2, f (x) is extremely stable (6 points) (II) when a ＜ 0, when f ′ (x) ＜ 0, we get x ＞ 0 or X ＜ - 2; when f ′ (x) ＞ 0, we get - 2 ＜ x ＜ 0; the monotone increasing interval of F (x) is (- 2,0); the monotone decreasing interval of F (x) is (- ∞, - 2) and (0, + ∞). (5 points) therefore, when x = 0, f (x) has a maximum, and its maximum is f (0) = 0 (6 points)
Given the function f (x) = TaNx, X ∈ (0,90 °), if x1, X2 ∈ (0,90 °), and x1 ≠ X2, it is proved that 0.5 [f (x1) + F (x2)] > F [(x1 + x2) / 2]
This is to prove that the image of tangent function is concave downward
f'(x)=1/(cosx)^2
f''(x)=2sinx/(cosx)^3>0
Therefore, it has been proved
If you are a senior high school student, you must use the definition to make a difference comparison
The function f (x) = LNX / x + 1 / X is known. When x > = 1, the inequality f (x) > = K / (x + 1) is always true
The known function f (x) = LNX / x + 1 / X
When x > = 1, the inequality f (x) > = K / (x + 1) holds, and the value range of real number k is obtained
F (x) > = K / (x + 1) is transformed into f (x) (x + 1) > = K
Let: (x + 1) (LNX / x + 1 / x) = LNX + 1 + LNX / x + 1 / X be g (x)
Derivation: get 1 / x + (1-lnx) / X & # 178; - 1 / X & # 178; = (x-lnx) / X & # 178; set the molecule H (x) because the denominator is greater than 0, so discuss the positive and negative of the molecule
The derivation is: 1-1 / X ∵ x > = 1 ∵ 1-1 / x > 0
The derivative of H (1) = 1 ∥ g (x) is always greater than 0 ∥ g (x)
So the minimum value of G (x) is g (1) = 2
∴k≥2
Given the function f (x) = LNX (1), find the maximum value (2) of the function g (x) = f (x + 1) - X. if for any x > 0, the inequality f (x) ≤ ax ≤ x ^ 2 + 1 holds,
Finding the value range of real number a
The derivative of the solution 1 function g (x) = f (x + 1) - x = ln (x + 1) - x (x ＞ - 1) gives G '(x) = 1 / (x + 1) - x = (1-x (x + 1)) / (x + 1) = (- x ^ 2-x + 1) / (x + 1) Let G' (x) = 0, that is - x ^ 2-x + 1 = 0, that is, x ^ 2 + X-1 = 0. The solution is x = (- 1 + √ 5) / 2 or X = (- 1 - √ 5) / 2. Therefore, when x belongs to (- 1, (- 1 + √ 5) / 2), G '(x) > 0, then G (x)