Find the function y = Tan (3x - π 3) And the monotone interval of the function is pointed out

Find the function y = Tan (3x - π 3) And the monotone interval of the function is pointed out

From 3x - π 3 ≠ K π + π 2, K ∈ Z, the definition domain of X ≠ K π 3 + 5 π 18, K ∈ Z. the function y = Tan (3x - π 3) is {x ≠ K π 3 + 5 π 18, K ∈ Z}. The value range is: (- ∞, + ∞). From − π 2 + K π < 3x − π 3 < π 2 + K π, K ∈ Z, we get − π 18 + K π 3 < x < 5 π 18 + K π 3, K ∈ Z

If the domain of function f (x-1) is (- 2,2), what is the domain of function f (3x + 1)? Which one is right

The domain of function f (x-1) is (- 2,2)
That is, X ∈ (- 2, 2)
So x - 1 ∈ (- 3, 1)
Then 3x + 1 ∈ (- 3, 1)
I.e. x ∈ (- 4 / 3, 0)

The function f (x) = (x2-3x + 3) * ex is known, and its definition field is [- 2, t] (T > - 2). Let f (- 2) = m, f (T) = n Verification: for any t > - 2, there is always x0 ∈ (- 2, t), which satisfies f '(x0) / ex0 = 2 / 3 * (t-1) 2, and determines the number of such x0

f'(x0)/e^x0=x0^2-x0=(x0-1/2)^-1/4,
‡ - 2 ≤ x0 = 1 / 2 ± root sign [2 / 3 (t-1) ^ 2 + 1 / 4] ≤ T, after simplification,
When taking the positive, the left side is obviously true, and the right side makes g (T) = t (t-1) ^ 3 ≤ 2 / 3 obviously exist (t = 1)
When taking negative, the proof on the right is the same as above, and 1 + 8 / 3 (t-1) ^ 2 ≤ 25 on the left obviously exists, so there are two

Given the function f (x) = (x2-3x + 3) * ex, its definition domain is [- 2, t] (T > - 2), Let f (- 2) = m, f (T) = n, try to determine the value range of T, so that the function

(1)f′(x)＝(x2－3x＋3)·ex＋(2x－3)·ex＝x(x－1)·ex.
By F '(x) > 0 ⇒ x > 1 or X

Given that the domain of function f (x) is (a, b) and B-A > 2, the domain of F (x) = f (3x-1) - f (3x + 1) is __

∵ the definition field of F (x) is (a, b) and B-A > 2,
‡ f (x) = f (3x-1) - f (3x + 1) shall meet
a＜3x−1＜b
a＜3x+1＜b ，
Namely
a+1
3＜x＜b+1
three
a−1
3＜x＜b−1
three ；
∵b-a＞2，∴b＞a+2，
∴b−1
3＞a+2−1
3=a+1
3，
The value range of X is a + 1
3＜x＜b−1
3；
The definition field of F (x) is (a + 1)
3，b−1
3）．
So the answer is: (a + 1)
3，b−1
3）．

It is known that the definition domain of function f (x + 2) is (- 2,5) to find the definition domain of function f (3x-1)

(f) no matter what is in the bracket, the range in the bracket is certain
The definition domain only refers to the scope of X
Title Theory
The definition field of F (x + 2) is (- 2,5), that is - 2, but the parenthesis is x + 2, so we need to find the range of X + 2 first
Right, ① + 2, you get
-2 + 2 is 0, that is, the range of parentheses in F () is (0,7)
Then for f (3x-1), 3x-1 in brackets must also be within the range of (0,7)
That is, 0 < 3x-1 < 7
The solution is 1 / 3. This is the range of X, that is, the definition domain
Namely
Definition field of function f (3x-1) (1 / 3.8 / 3)
What else is not very clear
You can ask