Given the function f (x) = sin ^ 2x + sinxcosx + 2cos ^ 2x (1) find the maximum value and minimum positive period of the function; (2) find the range of X which is equal to 1.5

Given the function f (x) = sin ^ 2x + sinxcosx + 2cos ^ 2x (1) find the maximum value and minimum positive period of the function; (2) find the range of X which is equal to 1.5

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Find the monotone interval of the function f (x) = sin ^ 2x sinxcosx

f(x)=sin^2x-sinxcosx
= (1-cos2x)/2-(1/2)sin2x
=(-1/2)(sin2x+cos2x)+1/2
=-(√2/2)*[sin2x*(√2/2)+cos2x*(√2/2)]+1/2
=(-√2/2)*[sin2x*cos(π/4)+cos2x*cos(π/4)]+1/2
=(-√2/2)*sin(2x+π/4)+1/2
(1) The increasing interval is the decreasing interval of y = sin (2x + π / 4)
∴ 2kπ+π/2≤2x+π/4≤2kπ+3π/2,k∈Z
∴ 2kπ+π/4≤2x≤2kπ+5π/4,k∈Z
∴ kπ+π/8≤x≤kπ+5π/8,k∈Z
The increasing interval is [K π + π / 8, K π + 5 π / 8], K ∈ Z
(1) The decreasing interval is the increasing interval of y = sin (2x + π / 4)
∴ 2kπ-π/2≤2x+π/4≤2kπ+π/2,k∈Z
∴ 2kπ-3π/4≤2x≤2kπ+π/4,k∈Z
∴ kπ-3π/8≤x≤kπ+π/8,k∈Z
The minus interval is [K π - 3 π / 8, K π + π / 8], K ∈ Z

Find the minimum positive periodic monotone interval of the function y = sinxcosx + sin? X

y=1/2(2sinxcosx)+(1-cos2x)/2
=(sin2x)/2-(cos2x)/2+1/2
=(√2)/2 [(√2)/2 sin2x - (√2)/2 cos2x ] +1/2
=(√2)/2 [cos（π/4） sin2x - sin(π/4) cos2x ] +1/2
=(√2)/2 sin(2x-π/4)+1/2
So the minimum positive period T = 2 π / 2 = π
The monotone increasing interval is [- π / 8 + K π, 3 π / 8 + K π] (K ∈ n +)
The monotone decreasing interval is [3 π / 8 + K π, 7 π / 8 + K π] (K ∈ n +)

Find the maximum and minimum positive period of the function f (x) = sin ^ 2x + sinxcosx

f(x)=(1-cos2x)/2+(sin2x)/2
=(1/2)sin2x-(1/2)cos2x+1/2
=(√2/2)*[sin2x*cos(π/4)-cos(2x)sin(π/4)]+1/2
=(√2/2)sin(2x-π/4)+1/2
T=2π/2=π,
The maximum value is (√ 2 / 2) + (1 / 2)
The minimum value is - (√ 2 / 2) + (1 / 2)

Find the zeros of the following functions and the coordinates of the vertex of the image, draw the sketch of each function, and point out the interval in which the function value is greater than 0 and which interval is less than 0 （1）y=1/3x²-2x+1 (2)y=-2x²-4x+1

This paper investigates the related knowledge of quadratic function y = ax 2 + BX + C
1）y=1/3x²-2x+1 
The opening direction is upward,
The symmetry axis X = 3, f (3) = - 2, so the vertex coordinates are (3, - 2),
Using the root formula to solve 1 / 3x? - 2x + 1 = 0, x = 3 ± √ 6
Therefore, when x < 3 - √ 6 or x > √ 6, the function value is greater than 0;
When 3 - √ 6 < x < 3 + √ 6, the function value is less than 0
diagram
2. Similar to (1)

The value range of the function y = sin | x | + SiNx is

x> 0 y = 2sinx range - 2 to 2
X

The value range of the function y = SiNx sin | x | It's said that Moda is popular The value range of the function y = SiNx sin | x | The one on the top just now is pasted wrong

When x > 0, | x | = x, y = 0
When x < = 0, y = SiNx sin (- x) = 2sinx, the range is [- 2,2]
To sum up, the value range is [- 2,2]

In order to get the image of the function y = sin (x + 8 times π), we need to make the image of function y = SiNx have a process

Left plus right minus
So it's eight tenths to the left

How can the image of the function y = SiNx / 2 be transformed by the function y = sin (x / 2 + π / 3) A. Shift π / 3 B to the left, π / 3 C to the right, 2 π / 3 B to the left, 2 π / 3 to the right But how am I supposed to be d?

Look at the title clearly! Y = sin (x / 2 + π / 3) becomes y = SiNx / 2
Not y = SiNx / 2 becomes y = sin (x / 2 + π / 3)!
Orz, my deskmate and I made the same mistake, the title of this question... Alas

Make the function y = SiNx + sin|x|, X belongs to the image of R

When x > = 0
f(x)=sinx+sin|x|=2sinx
When x