When x ∈ [0,1], find the minimum value of the function f (x) = x & sup2; + (2-6a) x + 3A & sup2

When x ∈ [0,1], find the minimum value of the function f (x) = x & sup2; + (2-6a) x + 3A & sup2

Axis of symmetry x = 3a-1 (sketch quadratic function)
① When 3a-1 ≤ 0, i.e. a ≤ 1 / 3,
The minimum value is obtained at x = 0, f (0) = 3A & # 178;
② When 0

Find the minimum value of F (x) = x & sup2; + (2-6a) x + 3A when x ∈ [0,1]

Function f (x) = x ^ 2 + (2-6a) x + 3A
With the opening upward, the axis of symmetry x = - (2-6a) / 2 = 3a-1
If 3a-1 ≤ 0, that is: a ≤ 1 / 3
The function f (x) = x ^ 2 + (2-6a) x + 3A increases monotonically on [0,1]
When x = 0, the minimum f (0) = 0 ^ 2 + (2-6a) * 0 + 3A = 3A
If 0

Let y = sin (Paix / 2 + Pai / 3), if for any x ∈ R, there exists x1, X2 such that f (x1)

It can be seen from the meaning that f (x1) = f (x) min = - 1 = > sin (π / 2x1 + π / 3) = - 1 = > π / 2x1 + π / 3 = 2K1 π - π / 2 = > X1 = 1 / (4k1-5 / 3) similarly, f (x2) = f (x) max = 1 = > sin (π / 2x2 + π / 3) = 1 = > π / 2x2 + π / 3 = 2k2 π + π / 2 = > x2 = 1 / (4k2 + 1 / 3) | x1-x2 | = | 1 / (4k1-5 / 3) - 1 / (4k2 + 1 / 3) |

Given the function f (x) = sin (π / 2 * x + π / 5), if f (x1) ≤ f (x) ≤ f (x2) holds for any x ∈ R, then the minimum value of | x1-x2 | is obtained

If f (x1) ≤ f (x) ≤ f (x2) holds for any x ∈ R, then - 1 ≤ f (x) ≤ 1,
The shortest distance between the minimum and the maximum is half a period,
The period T = 2 π / (π / 2) = 4, and the minimum value of | x1-x2 | is 2

The function f (x) = 2cosx, for X belongs to R, f (x1) is less than or equal to f (x) is less than or equal to f (x2), then the minimum absolute value of x1-x2
The function f (x) = 2cosx, for X belongs to R, f (x1) is less than or equal to f (x) is less than or equal to f (x2), then what is the minimum absolute value of x1-x2

The minimum absolute value of X 1-x 2 is π

The function f (x) = 3sin (2x - π / 8), for any x belonging to R, f (x1) ≤ f (x) ≤ f (x2), then the minimum absolute value of x1-x2 is?

For any x belonging to R, f (x1) ≤ f (x) ≤ f (x2),
So f (x2) is the maximum of F (x) and f (x1) is the minimum of F (x),
The minimum value of the absolute value of x1-x2 is f (x), and the half period is π / 2

Given the function f (x) = x2 + 2x + A / x, X ∈ [1, + ∞), when a = - 1, find the minimum value of function f (x)
The key will be ideas or methods,

Given the function f (x) = (X & # 178; + 2x + a) / x, X ∈ [1, + ∞), when a = - 1, find the minimum value of function f (x). When a = - 1, f (x) = (X & # 178; + 2x-1) / x = x - (1 / x) + 2, because f '(x) = 1 + (1 / X & # 178;) = (X & # 178; + 1) / X & # 178; > 0 holds for any x, so f (x) is an increasing function in its domain, then

Given the function f (x) = x2 + 2x + 3 (x ∈ [2, + ∞)), find the minimum value of F (x)?

f(x)=(x+1)^2+2
So the minimum value is f (2) = 4 + 4 + 3 = 11

Given that the function f (x) = (13) x, X ∈ [- 1, 1], the minimum value of function g (x) = f 2 (x) - 2 AF (x) + 3 is h (a), we can find H (a)

Let t = (13) x, t ∈ [13, 3]. & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; then G (x) min = H (a) = φ (13) = 289-2a3 when a < 13; G (x) min = h (a) = 3-a2 when 13 ≤ a ≤ 3; G (x) min = H (a) = 3-a2 when a > 3

If the minimum value of function FX = | x | + | 2x + a | is 3, then the value of real number a is 3

A = 6 or - 6
f(x)=|x-0|+2|x-(-a/2)|
The minimum value from a point on the number axis to 0 and - A / 2 is 3
Then this point is between 0 and - A / 2, and the distance between 0 and - A / 2 is the minimum