Find the period of given function y = cosx [cosx cos (x + π / 3)]

Find the period of given function y = cosx [cosx cos (x + π / 3)]

Y = cosx [cosx-1 / 2cosx + 3 ^ 2 / 2sinx] = 1 / 2 cos ^ 2 + 3 ^ 2 / 2cosxsinx = (1-cos2x) / 4 + 3 ^ 2sin2x / 4 = 1 / 4-cos2x / 4 + 3 ^ 2sin2x / 4 = 1 / 4-1 / 2 [cos2x / 2-3 ^ 2sin2x / 2] = 1 / 4-1 / 2cos (2x + 60 degrees)
T = 2 faction
Please put the root of 3 ^ 2 into three
Is the relationship between cosx and the things in the brackets multiplied or not?
Good morning, I forgot all the things that my teacher taught. I wish you progress in your study and success in the college
Find the range of function f (x) = x ^ 2-4x + 1 (x ≥ a)
f(x)=(x-2)^2-3
The opening is upward and the axis of symmetry is x = 2
Discuss a separately
1) If a > = 2, then Fmin = f (a) = a ^ 2-4a + 1, the range is [a ^ 2-4a + 1, + ∞)
2) If a
The specified sign "△" denotes an operation, a △ B = a - √ (AB) + 2B. If 1 △ k = 0, then the range of function f (x) = k △ (4x / 9) is
Sorry, wrong number. It's 1 △ k = 4
1△k=1-√k+2k=4;
2(√k)^2-√k-3=0;
(√k+1)·(2√k-3)=0;
∵√k≥0,∴√k+1>0.
∴2√k-3=0.
Then: √ k = 3 / 2;
k=9/4;
■: function f (x) = (9 / 4) △ (4 / 9) · x
=(9/4)-√[(9/4)·(4/9)x]+2·(4/9)·x
=(9/4)-√x +8x/9
=(8/9)[(√x)^2 -(9/8)·√x+81/32]
=(8/9)[(√x - 9/16)^2 + 81×7/256]
≥(8/9)×(0+ 81×7/256)
=63/32.
The range is [63 / 32, + ∞)
.
1 △ k = 1 - √ K + 2K = 0, K has no solution
Given the parabola y = x & sup2; - (3m-1) x + 9m-1, no matter what the value of X is, the value of function y is non negative. How to find the value range of M?
From X to 0
Easy to get
As long as the equation x & sup2; - (3m-1) x + 9m-1 = 0
The △ of is less than 0
That is (3m-1) & sup2; - 4 (9m-1) < 0
The solution is its value range
The value of the function y is nonnegative, indicating that the vertex of the parabola is on or above the X axis
Then the ordinate of the vertex is ≥ 0
According to the vertex formula (4ac-b & sup2;) / 4A = [4 (9m-1) - (3m-1) & sup2;] / 4 ≥ 0
Then (7-2 √ 11) / 3 ≤ x ≤ (7 + 2 √ 11) / 3
y=x²-(3m-1)x+9m-1
Find the pole
dy/dx=0
It is concluded that x = (3m-1) / 2
y=[(3m-1)/2]²-(3m-1)(3m-1)/2+9m-1>=0
9m²-42m+5>=0
The results show that M is between 0.12225 and 4.54441
According to the meaning of the title,
The opening of function image is upward, and there is no intersection with X axis
M>0
△=4(3m-1)^2-4m(9m-1)1/5
M range: M > 1 / 5 helps me
Answer time: 23:08, September 9, 2010
Definition field and function value of F (x) = LG (2-x) of X-1
The domain must meet the following requirements:
2-x>0
x-1≠0
That is, the domain of definition is: X
Definition field: from X-1 is not equal to 0, X is not equal to 1
X is obtained from 2-x > 0
Let's know the parabola y = MX & sup2; - 2 (3m-1) + 9m-1, no matter what value x takes, y is a non negative number, and find the value range of M?
-2(3m-1)x
According to the meaning of the title
m> And 4 (3m-1) & sup2; - 4m (9m-1) ≤ 0
You can get it
M > 0 and 5m > 1
So m > 1 / 5
The domain of function f (x) = LG (x ^ 2-2x-3) is
LG [(x + 1) (x-3)], then (x + 1) (x-3) > 0, so x > 3 or X
Let f (x) = | lgx |, x > 0, f (x) = - x ^ 2-2x, X be a function of R
Wait tonight.
First of all, you need to know what is zero point? Zero point is the value of independent variable x when f (x) = 0. When x > 0, f (x) = | lgx | = 0, that is, lgx = 0, then x = 1. When x ≤ 0, f (x) = - x ^ 2-2x = 0, then x = 0 or x = - 2, so the function f (x) with defined field r has three zeros of - 2,0,1
Definition field of function f (x) = 1 / X * LG (3 + x)
{x ﹥ x ﹥ 3, and X ≠ 0}
The domain of F (x) = 1 / [xlg (3 + x)]
It is (- 3, - 2) ∪ (- 2, 0) ∪ (0, + ∝)
Finding the number of zeros of function f (x) = lgx + 2x-7
There is only one zero
Let lgx + 2x-7 = 0  lgx = - 2x + 7 be the intersection of two functions
Draw the image of y = lgx and y = 7-2x in the coordinate plane
See how many intersections are zero points
Derivation
Monotone increasing function
When x = 1 - 5
13 when x = 10
So one zero
It can be seen as the number of image intersections of F (x) = LNX and G (x) = - 2x + 7
Then draw a picture
The intersection of F (x) monotonically increasing on X axis is (1,0)
The intersection (3.5,0) of G (x) monotone decreasing and X axis
So there is a focus between X ∈ (1,3.5)
So there's a zero
First of all, the original formula is 0, and the number of intersections of y = lgx and y = - 2x + 7 is obtained by shifting the term. Do the image, get an intersection, that is, the number of zeros is 1. Hope to adopt.
The number of zeros of F (x) = lgx + 2x-7 is equivalent to the number of intersections of G (x) = lgx and H (x) = - 2x + 7,
When x > 0, the function value increases from - ∞ to + ∞,
When x > 0, the function value decreases from 7 to - ∞,
So their images have only one intersection, so the number of zeros of F (x) = lgx + 2x-7 is 1. ... unfold
The number of zeros of F (x) = lgx + 2x-7 is equivalent to the number of intersections of G (x) = lgx and H (x) = - 2x + 7,
When x > 0, the function value increases from - ∞ to + ∞,
When x > 0, the function value decreases from 7 to - ∞,
So their images have only one intersection, so the number of zeros of F (x) = lgx + 2x-7 is 1. Put it away