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Given that the definition domain of function f (x) is [0,1], find the definition domain of function g (x) = f (x + m) + F (x-m) (M > 0)
∵ the definition field of function f (x) is [0, 2], ∵ to make the function g (x) = f (x + m) + F (x-m) (M > 0), then 0 ≤ x + m ≤ 20 ≤ x − m ≤ 2, i.e. − m ≤ x ≤ 2 − mm ≤ x ≤ 2 + m, ∵ m > 0, ∵ when 2-m = m, i.e. M = 22, then x = 22, if 0 < m < 22, then m ≤ x ≤ 2-m, if M > 2
If the domain of function f (x) is [- 2,1], find the domain of G (x) = f (x) + F (- x)
The definition field of F (x) is [- 2,1] [that is, f can process numbers between - 2 and 1]
The definition field of F (- x) is [- 1,2] [that is, - x is the number between - 2 and 1]
The definition field of G (x) = f (x) + F (- x) is [- 1,1] [i.e. x [∵ f (x)] should be between - 2 and 1, and also between (∵ f (- x)) [- 1,2]
Given the definition domain [0,1] of function f (x), find the definition domain of G (x) = f (x + m) + F (x-m) Column inequality group: 0=
-m<=x<=1-m [when m ＞ 1 / 2, X ≤ 1-m ＜ 1 / 2]
m<=x<=1+m [when m > 1 / 2, X ≥ m > 1 / 2]
So, inequality group
0=
Therefore, when m > 1 / 2, the definition field is an empty set
Known function f (x) = 3x, f (a + 2) = 18, G (x)= λ• The semantic domain of 3ax-4x is [0, 1] (I) find the value of a; (II) if the function g (x) is a monotonically decreasing function on the interval [0,1], find the real number λ Value range of
(I) 3A + 2 = 18 ⇒ 3A = 2 ⇒ a = log32
(II) g (x)= λ• 2x-4x
Let 0 ≤ X1 < x2 ≤ 1, because g (x) is a monotonic decreasing function on the interval [0,1]
So g (x1) - G (x2) = (2x2-2x1) (- λ+ 2x2 + 2x1) ≥ 0
∵2x2-2x1＞0
∴ λ ≤ 2x2 + 2x1 constant, because 2x2 + 2x1 ≥ 20 + 20 = 2
So real numbers λ The value range of is λ ≤2
Find the generalized integral ∫ Xe ^ (- x ^ 2) DX, where the upper limit of the integral is + ∞, and the lower limit of the integral is 1,
Approximate differential
∫xe^(-x^2)dx
=∫ e ^ (- x ^ 2) d (x ^ 2 / 2) because xdx = D (x ^ 2 / 2)
=-1/2*∫e^(-x^2)d(-x^2)
=-1/2*e^(-x^2)
-1 / 2 * e ^ (- x ^ 2) is 0 at infinity
Where x = 1 is - 1 / 2 * e ^ (- 1)
So the result is 0 - [- 1 / 2 * e ^ (- 1)] = 1 / 2 * e ^ (- 1)
Integral problem ∫ [2 ~ 0] (4x ^ 3-x ^ 4) ^ (1 / 2) DX how to calculate
∫ (4x ^ 3-x ^ 4) ^ (1 / 2) DX integral limit 2,0 = ∫ x [(4x-x ^ 2) ^ (1 / 2)] DX = ∫ x [(4 - (X-2) ^ 2] ^ (1 / 2)] dxx-2 = t = ∫ (T + 2) [(4-T ^ 2] ^ (1 / 2)] DT integral limit 0, - 2 = ∫ t [(4-T ^ 2] ^ (1 / 2)] DT + ∫ 2 [(4-T ^ 2] ^ (1 / 2)] DT integral limit 0, - 2 the first part is easy to find, the second part is troublesome, use t
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