![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
Let f (x) be an even function defined on R and a decreasing function on [0, + 8]. Try to compare (- 3 / 4) with F (a2-a + 1)
a^2-a+1=(a-1/2)^2+3/4>=3/4
f(-3/4)=f(3/4)
[0, + 8) is a decreasing function, so
f(-3/4)
Given that the function f (x) is a decreasing function in the interval (0, positive infinity), then the relationship between F (a ^ 2 + A + 1) and f (3 / 4) is
Solution a ^ 2 + A + 1-3 / 4
=a^2+a+1/4
=(a+1/2)^2
≥0
That is, a ^ 2 + A + 1 ≥ 3 / 4
From the known function f (x) is a decreasing function in the interval (0, positive infinity)
Know f (a ^ 2 + A + 1) ≤ f (3 / 4)
If the function f (x) = x & # 178; + BX + C satisfies: F (1) ∈ [- 3,2], f (2) ∈ [1,8], then the value range of F (3) is
If f (1) ∈ [- 3,2], we can get: - 3 ≤ 1 + B + C ≤ 2.1
If f (2) ∈ [1,8], we can get: 1 ≤ 4 + 2B + C ≤ 8.2
2 Formula x2 + 1 formula x (- 1)
0 ≤ 7 + 3B + C ≤ 19, that is: - 7 ≤ 3B + C ≤ 12.3
Because: F (3) = 9 + 3B + C, we can get: 3B + C = f (3) - 9
-7≤f(3)-9≤12
The solution is: 2 ≤ f (3) ≤ 21
When the function y = x ^ 2 + BX + C (x ∈ (- ∞, 1)) is monotone, the value range of B
A quadratic function is monotone in an interval
The axis of symmetry is no longer in this interval, but can be on the boundary
The axis of symmetry of Y is x = - B / 2
So - B / 2 is not in (- ∞, 1)
So - B / 2 ≥ - 1
b≤-2
The function f (x) = x & # 178; + BX + 2 is known. If f (x) ≥ B + 3 is constant when x ∈ [- 1,4], then f (x) is obtained
f(x)=x²-2x+2 （b=-2）
In this paper, the distribution of the minimum value is discussed with the idea of discussing the symmetry axis by classification, and compared with the given conditions, the value of B is finally obtained
f(x)=x²+bx+2=f(x)=（x+b/2)²+2-b²/4
The opening of the quadratic function is upward, the axis of symmetry is x = - B / 2, and the vertex is (- B / 2,2-b & # / 4)
1. If the axis of symmetry is less than or equal to - 1
-B / 2 ≤ - 1 (i.e. B ≥ 2),
The minimum value of F (x) is f (- 1),
There must be f (- 1) = 1-B + 2 ≥ B + 3 to get B ≤ 0
2. If the axis of symmetry is in the interval [- 1,4], (- 8)
Given the function f (x) = log3x + 2 (x ∈ [1,9]), find the maximum value of y = [f (x)] &# 178
X ∈ [1,9], so 0 ≤ log3x ≤ 2, so 2 ≤ f (x) ≤ 4, then the maximum value of Y is 4 ^ 2 = 16
Given m ∈ R, the function f (x) = (x2 + MX + m) ex. (1) if the function f (x) has no zero point, find the value range of real number m; (2) if the function f (x) has a maximum value and is written as G (m), find the expression of G (m); (3) when m = 0, find out: F (x) ≥ x2 + X3
(1) Let f (x) = 0, get (x2 + MX + m) · ex = 0, so x2 + MX + M = 0. Because f (x) has no zero point, so △ = m2-4m < 0, so 0 < m < 4. (4 points) (2) f '(x) = (2x + m) ex + (x2 + MX + m) ex = (x + 2) (x + m) ex, Let f' (x) = 0, get x = - 2, or x = - m, when m > 2, - M < - 2. List the following table: X (- ∞, - M) - M (- m, - 2) - 2 （-2，+∞） f'（x） + 0 - 0 + f（x） ↗ me-m ↘ （4-m）e-2 ↗ When x = - m, f (x) has a maximum value me-m. (6 points) when m = 2, f '(x) = (x + 2) 2ex ≥ 0, f (x) is an increasing function on R, so f (x) has no maximum value. (7 points) when m < 2, - M > - 2. List the following table: X (- ∞, - 2, - M) - M (- m, + ∞) f' (x) + 0 - 0 + F (x) ↗ （4-m）e-2 ↘ me-m ↗ When x = - 2, f (x) gets the maximum value (4-m) E-2, (9 points) so g (m) = me-m, M > 2 (4-m) E-2, m < 2 (10 points) (3) when m = 0, f (x) = x2ex, let ϕ (x) = ex-1-x, then ϕ '(x) = EX-1, when x > 0, φ' (x) > 0, and φ (x) is an increasing function; when x < 0, φ '(x) < 0, and φ (x) is a decreasing function, so when x = 0, the minimum value of φ (x) is 0. (1) Therefore, f (x) ≥ x2 + X3. (16 points)
Given the function f (x) = x square + 2mx + m square-m / 2-3 / 2, when x ∈ (0, + ∞), f (x) > 0, find the range of M
Such as the title
-When m ≤ 0, i.e. m ≥ 0, f (0) minimum = m ^ 2-0.5m-1.5 > 0 m > 0.5 - M > 0 M0 M
It is known that f (x) is a decreasing function defined on (- 1,1), and f (1-A)
1-A, a & # - 1 are all in the domain
-1
Given that f (x) = x & # 178; + 4ax + 2 is a decreasing function in (- ∞, 6), then the value range of F (1) is obtained
F '(x) = 2x + 4a, 2x + 4A = 0, the solution is x = - 2A, because the function is a decreasing function in (- ∞, 6), so the solution of - 2A > = 6 is a
RELATED INFORMATIONS
- 1. If the quadratic function f (x) = x ^ 2 - (A-1) x + 5 is monotone in the interval (0,1), then the value range of a is
- 2. F (x) = (m-1) x ^ 2 + (m-2) x + (m ^ 2-7m + 12), X ∈ [3-A, 5] is even function, find the value of a, M
- 3. Given that in a function y = (1 + 2m) x, the value of function y decreases with the increase of independent variable x, then the value range of M is () A. m≤−12B. m≥−12C. m＜−12D. m＞−12
- 4. If the square (x2-m) of the function f (x) = (x-1) has a maximum at x = 1, then the value range of M is
- 5. Try to find out an interval of length 1, in which the function X-1 / 3x + 2 has at least one zero operation problem It seems that the substitution of x = 1 is not correct, as if it does not satisfy f (a) f (b) of finding zero
- 6. The minimum and maximum of function f (x) = x2 ax (a) in the interval [0.1]
- 7. F (x) = 1 / 3x3 ln (2 + x3), find the maximum and minimum of the function
- 8. The number of intersections of function f (x) = x ^ 3-6x ^ 2 + 9x-10 and y = - 8 is
- 9. The sufficient and necessary condition is that the maximum point of continuous function f (x) on interval [a, b] is the point where the function obtains the maximum value in this interval? In addition, if it is changed to the minimum value point and minimum value, the conclusion should be the same? When learning the binomial theorem, there is a kind of problem which is to find the term with the largest coefficient. Let PN denote the coefficient, then the standard solution is the n-value satisfying the equations PN > PN + 1, PN > PN-1. But the N solved by this formula represents the maximum point of function f (n) = PN, The maximum is not necessarily the maximum, but the standard answer here directly regards the maximum as the maximum. When are the two equal?
- 10. Given the function f (x) = x2 + 2x-3, the set M = {(x, y) | f (x) + F (y) ≤ 0}, the set n = {(x, y) | f (x) - f (y) ≥ 0} Finding the intersection of M and n How to make f (x) - f (y) = 0?
- 11. When k is the value, the domain of the function y = LG (KX ^ 2 + 4kx + 3) is r? When k is the value, the domain is r?
- 12. It is known that the functions f (x) = asin (KX + π / 3) and φ (x) = btan (KX - π / 3), k > 0, and a; B belong to R. the sum of the minimum positive periods of the two functions is 3 π / 2, and f (π / 2) = φ (π / 2), f (π / 4) = - √ 3 φ (π / 4) + 1, find the analytic expressions of two functions
- 13. The known function f (x) = ax & # 178; - x + 2a-1 If a > 0, let g (a) be the minimum value of F (x) in the interval [1.2]
- 14. Why to find the increasing interval of logarithmic function is to find the decreasing interval of compound function
- 15. Given the function f (x) = AX2 + BX + C, f (- 3) = f (1) = 0, f (0) = - 3, we can find the solution set of the equation f (x) = 2x
- 16. Given the set a = {(x, y) y = f (x), X belongs to [a, b]}, B = {(x, y) x = 1}, then the number of elements contained in a intersection B is?
- 17. Let a = {(x, y) / y = x ^ 2}, B = {(x, y) / y = 2 ^ x}, then the number of subsets a intersects B ()
- 18. The set of positive integers divided by 5 and 2 is expressed as
- 19. Given that P is equal to {0,1}, P is a set or an element
- 20. Set a = {x | X & sup2; - 3x + 2 = 0}, B = {x | X & sup2; + 2 (a + 1) x + A & sup2; - 5 = 0} (1) If a ∩ B = {2}, find the value of A (2) If a ∪ B = a, find the value range of A (3) If u = R, a ∩ (cub) = a, find the value range of real number a