Find the value range of (x's square + 5x) - 1 under the root sign of function y = 5LN

Find the value range of (x's square + 5x) - 1 under the root sign of function y = 5LN

y = 5ln(x²+5x) - 1
∵ zero and negative numbers have no logarithm
∴x²+5x=x(x+5)＞0
The definition field x ＜ - 5, or X ＞ 0
∵ x ∵ x ∵ x ᙽ 5x = (x + 5 / 2) 2 ∵ 25 / 4 can take all positive numbers
∴5ln(x²+5x)∈R
∴5ln(x²+5x) - 1 ∈R
The value range is r

1.y=x.(6-3x) (0

Y = x · (6-3x) = - 3 · (x2-2x + 1) + 3 = - 3 (x-1) 2 + 3 {is the formula}
When x = 1, the function decreases. When x = 0, y = 0. When x = 2, y = 0. Therefore, y = X. (6-3x) (0)

Y = open radical (- xsquare - 2x + 3) The definition of the domain to find out how to calculate the value range

Put the second three formula in the root number:
-x^2-2x+3
=-(x^2+2x-3)
=-(x^2+2x+1)+4
=-(x+1)^2+4
It can be seen that the value range of S = - x ^ 2-2x + 3 is s

Find the value range of y = square of root - X - 2x + 3

Let m = - x? - 2 x + 3, then y = √ M
∵ m = - x ² - 2 x + 3
= -（x ² + 2 x ）+ 3
= - （x ² + 2 x + 1）+ 3 + 1
= -（x + 1）² + 4
When x = - 1, M has a maximum value of 4
∵ y = √m
∴ 0 ≤ m
∴ 0 ≤ m ≤ 4
∴ 0 ≤ √m ≤ 2
The value range is: [0,2]

Y = the value range of (- x squared + 2x + 3) in the root sign

【1】 When x ∈ [- 1,3], the function definition domain is [- 1,3]. [2] when x ∈ [- 1,3] - 4 - (x-1) Ω, it is easy to know that 0 ≤ - x 2 + 2x + 3 ≤ 4. = = > 0 ≤ y ≤ 2

What is the range of - x squared + 2x-3 under the root sign

-x^2+2x-3≥0
X^2-2X+3≤0
Δ = 4-3 * 1 * 4 less than 0
The solution set is empty

The minimum positive period, range and monotone interval of F (x) = sin2x + 2 radical 3cos2x

f(x)=sin2x+2√3cos2x=√(1+12)*sin(2x+γ),sinγ=2√39/13,γ=arcsin2√39/13
T = 2 π / 2 = π, range [- √ 13, √ 13]
Monotone interval is also made in turn

The minimum positive period of y = sin2x + radical 3cos2x-1 is ()

y=2(sin2x*1/2+cos2x*√3/2)-1
=2(sin2xcosπ/3+cos2xsinπ/3)-1
=2sin(2x+π/3)-1
So t = 2 π / 2 = π

Find the minimum positive period and maximum value of the function y = sin2x + radical 3cos2x Thank you very much

The original formula: y = sin2x + √ 3cos2x / y = 2 [(1 / 2) sin2x + (√ 3 / 2) cos2x]] = 2 [sin2xcos (π / 3) + cos2xsin (π / 3)]. = 2Sin (2x + π / 3). The minimum positive period T = 2 π / 2 = π. When 2x + π / 3 = 2K π + π / 2 = π, when 2x + π / 3 = 2K π + π / 2, that is x = k π + π / 12, K ∈ Z, sin (2x + π / 3) = 1, function y y, function y, function y y = 1, function y y y, function y y = 1, function y (2x + π / 3) = 1, it has a maximum value, ymax = 2

The known function f (x) = sin2x+ 3cos2x (1) Find the maximum and minimum value of function f (x); (2) Find the minimum positive period of function f (x); (3) Find the monotone increasing interval of function f (x)

（1）∵f(x)＝sin2x+
3cos2x=2sin(2x+π
3)，
So [f (x)] max = 2, [f (x)] min = 2
(2) The minimum positive period of the function is t = 2 π
2=π．
(3) Let 2K π - π
2≤2x+π
3≤2kπ+π
2, K ∈ Z, find K π − 5 π
12≤x≤kπ+π
12  ， (k∈Z)，
So the monotone increasing interval of the function is [K π − 5 π
12，kπ+π
12](k∈Z)．