According to the general steps of drawing function image, draw the image of function y = x + 1, and answer according to the image: (1) when the value of X is, the value of Y is 0; (2) when the value of Y is, the value of X is 0; (3) when the value of X is, Y > 0; (4) when the value of X is, y increases with the increase of X

Let x = 0, then y = 1; let y = 0, then x = - 1, so the image of linear function y = x + 1 passes through points (0, 1), (- 1, 0). The image is as shown in the figure: (1) as shown in the figure, when x = - 1, y = 0. (2) as shown in the figure, when y = 1, x = 0; (3) as shown in the figure, when x ＞ - 1, y ＞ 0; (4) as shown in the figure, when x is any real number, y increases with the increase of X
The general steps to draw a function image are
1. View domain
2. Observe the monotonicity of function
Find the inflection point (3
4. Draw the coordinate axis and trace the points
5. Outline the general picture
6. Complete the image and mark the coordinates of special points
F (x) = - 2x & # 178; - MX-3 if x ∈ [- 2, + ∞) is an increasing function and X ∈ (- ∞, - 2] is a decreasing function, then f (1)
∵ f (x) when x ∈ [- 2, + ∞) is an increasing function and X ∈ (- ∞, - 2] is a decreasing function
The axis of symmetry of F (x) is x = - 2
∴-m/4=-2
∴m=8
∴f(x)=-2x²-8x-3
∴f(1)=-2-8-3=-13
That is, x = - 2 is the axis of symmetry,
So - M / 4 = - 2
The result is: M = 8
f(1)=-2-m-3=-5-8=-13
f(1)=-13.
Because f (x) is an increasing function in X ∈ [- 2, + ∞), and X ∈ (- ∞, - 2] is a decreasing function, so x = - 2 is the inflection point of F (x), so the first derivative at this point is 0. The first derivative of f (x) is - 4x-m. take x = - 2 into - 4x-m = 0, and M = 8, so f (1) = - 2-8-3 = - 13
Find the value range of the following function y = root sign (2Sin ^ 2x + 3cosx-3)
y=√[2sin^2(x)+3cosx-3]
=√[2(1-cos^2(x))+3cosx-3]
=√[-2cos^2(x)+3cosx-1]
=√[-2[cosx-3/4]^2+1/8]
also
-2cos^2(x)+3cosx-1>=0
2cos^2(x)-3cosx+1
Sin ^ 2x is replaced by 1-cos ^ 2x, and then all cosx in the root sign are replaced by T. the definition field of T is less than + 1 and greater than - 1
In the root sign, the whole formula is regarded as a quadratic equation of one variable. It is quite simple to find the range of quadratic equation of one variable. Just look at the image of parabola
It is known that the function y = (M + 3) x to the power of 2m + 1 + 3 is a linear function
(1) Finding the value of M
(2) Draw the function image
To have a process and diagram, thank you very much!
Process, process! thank you
Because y = (M + 3) the 2m + 1 power of X + 3 is a linear function
So m + 3 is not equal to 0, 2m + 1 = 1
So m = 0
So y = 3x + 3
When x = 0, y = 3, when x = 1, y = 6
The graph is a straight line passing through (0,3), (1,6), draw it by yourself
Or when x = 0, y = 3, when y = 0, x = - 1
This is a better way to draw
Find the definition field and value field of y = root sign (2Sin ^ 2 (x) + 3cosx-3)
Such as the title
Solution:
because:
Y
=√[2sin^2(x)+3cosx-3]
=√[2(1-cos^2(x))+3cosx-3]
=√[-2cos^2(x)+3cosx-1]
Because the number of square root is not negative
Then:
-2cos^2(x)+3cosx-1>=0
2cos^2(x)-3cosx+1
∵y=√(2sin²x+3cosx-3)=√(-2cos²x+3cosx-1)。
If y has a solution in the range of real number, then - 2cos & sup2x + 3cosx-1 ≥ 0, that is, 2cos & sup2x-3cosx + 1 ≤ 0.
∴0≥2cos²x-3cosx+1=2(cosx-3/4)²-1/8；
∴1/16≥(cosx-3/4)²≥0；
That is: - 1 / 4 ≤ co... expansion
∵y=√(2sin²x+3cosx-3)=√(-2cos²x+3cosx-1)。
If y has a solution in the range of real number, then - 2cos & sup2x + 3cosx-1 ≥ 0, that is, 2cos & sup2x-3cosx + 1 ≤ 0.
∴0≥2cos²x-3cosx+1=2(cosx-3/4)²-1/8；
∴1/16≥(cosx-3/4)²≥0；
That is: - 1 / 4 ≤ cosx-3 / 4 ≤ 1 / 4, that is, 1 / 2 ≤ cosx ≤ 1.
∴x∈[π/3+2nπ，π/2+2nπ]，n∈Z。
∴y=√(2sin²x+3cosx-3)=√(-2cos²x+3cosx-1)=√[1/8-2(cosx-3/4)²]≤√(1/8)=(√2)/4
That is, y ∈ [0, (√ 2) / 4]
2sin^2（x）+3cosx-3>=0
2cos^2(x)-3cos(x)+1
F (x) is an even function defined on R, increasing from 0 to positive infinity, and f (1 / 2) = 0, which solves the inequality f (lgx) > 0
Because f (x) is an even function defined on R, increasing from 0 to positive infinity, and f (1 / 2) = 0, then f (- 1 / 2) = 0, let lgx = t, x > 0, then f (T) = f (lgx) > 0, t > 1 / 2 or T1 / 2 or lgx0), x > 10 under the root, or 0
Given the function FX = √ 3 (sin ^ 2x cos ^ 2x) - 2sinxcos 1. Find the minimum positive period of FX
2. Let x ∈ [- π / 3, π / 3] find the range and monotone increasing interval of FX
FX = - √ 3cos2x-sin2x = - 2Sin (2x + π / 3), so the minimum positive period is π
F'x = - 4cos (2x + π / 3), when f'x > 0, the increment of X is on (π / 12, π / 3), f'x = 0, x = π / 12. The minimum f (π / 12) = - 2
F (- π / 3) = √ 3. F (π / 3) = 0. The value range of F (x) is [- 2, radical 3]
It is defined that the even function f (x) on R monotonically decreases on the interval [0, positive infinity], if f (1)
Because the even function, [0, positive infinity) is decreasing again, it is easy to get
-1
Let f (y) = (x + F) ∞ be defined as (x + y), and let f (y) = (x + F) ∞ be (x + y)
Let y = x have: F (2x) + F (0) = 2F (x) f (x)
Let y = - X have: F (0) + F (2x) = 2F (x) f (- x)
So 2F (x) f (x) = 2F (x) f (- x)
So even function is f (x) = x (f)

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