What is the function of y = cos (π / 4 + x) with respect to origin symmetry?

What is the function of y = cos (π / 4 + x) with respect to origin symmetry?

Trigonometric function is also a function, and - f (- x) and f (x) are symmetric about the origin! So y = - cos (π / 4-x)
Judging the symmetry of the domain about the origin
Is domain [- 5,3) symmetric about origin?
Is it true that if x is not equal to a value and X contains the opposite number of this value, then the domain of X must not be symmetric about the origin?
For example: F (x) = 1 + SiNx cosx / 1 + SiNx + cosx
Because the denominator is not zero, {x | x is not equal to 2K π - π / 2 and X is not equal to 2K π + π}
Why is the domain of definition not symmetrical about the origin?
Asymmetry, draw the number axis, you can see it at a glance
Is it true that if x is not equal to a value and X contains the opposite number of this value, then the domain of X must not be symmetric about the origin?
yes
If the set a = {x x & # 178; - 2x + m} contains two elements, what are the conditions for the real number m to satisfy
Then the equation of X has two real solutions
So △ = B & # 178; - 4ac > 0
(-2)²-4m>0
4>4m
So m
It is known that x = 1 is an extreme point of the function f (x) = MX ^ 3-3 (M + 1) x ^ 2 + NX + 1, where m and N belong to R, M
f'(x)=3mx^2-3(m+1)x+n,f'(1)=0,3m-n+6=0,
f'(x)=3mx^2-6(m+1)x+3m+6,x1=m,x2=1+2/m,m
Wrong upstairs
Let X and y be real numbers, and analyze the following four sets: a = {y y = x & # 178; + 1} B = {xy = x & # 178; + 1}
C = {y y = x & # 178; + 1} D (x, y y = x & # 178; + 1} do they have the same meaning? Why?
C = {y y = x & # 178; + 1} D (x, y y = x & # 178; + 1} they have different meanings,
C = {y y = x & # 178; + 1} denotes the set of function values, C = {y| y} 1}
D {(x, y) y = x & # 178; + 1} denotes the set of all points on a parabola
It is known that x = 1 is an extreme point of the function f (x) = MX ^ 3 - 3 (M + 1) x ^ 2 + NX + 1, where m and N belong to R, m ≠ 0. Find the relationship between M and N, and find the interval of FX
f'(x)=3mx^2-6(m+1)x+n
F '(1) = 0 = 3m-6 (M + 1) + n = - 3m-6 + n
That is: 3m-n + 6 = 0
The set of all real number solutions of the equation x & # 178; + X + 1 = 0 is expressed by description
Because the equation has no solution, it is an empty set
It is known that x = 1 is an extreme point of the function f (x) = mx3-3 (M + 1) x2 + NX + 1, where m, n ∈ R, m < 0 (1) find the relationship between M and n
(2) When x is greater than or equal to - 1 and less than or equal to 1, the tangent slope of any point on the image of function = f (x) is always greater than 3m, and the value range of M is obtained,
1、n=3m+6
2、f'(x)＝3mx^2-6(m+1)x+3m+6
=3m[x-(m+1)/m]^2-3/m
Because m3m, because f '(1) = 0, so 3m0 or m
That wonderful answer is wrong! / L /!!!!!!!
What about the set of real number solutions of equation x2 + X + 1 = 0?
Write out the elements in the above set respectively, and point out what set is in the above set
The discriminant of the equation is less than 0, so the equation has no real solution, and the elements in the set are two complex roots
It is known that x = 1 is an extreme point of the function f (x) = mx3-3 (M + 1) x2 + NX + 1, where m, n ∈ R, m ＜ 0. (I) find the expression of the relationship between M and N; (II) find the monotone interval of f (x); (III) when x ∈ [- 1, 1], the tangent slope of any point on the image of function y = f (x) is greater than 3m, and find the value range of M
(I) f '(x) = 3mx2-6 (M + 1) x + n. because x = 1 is an extreme point of F (x), f' (1) = 0, that is, 3m-6 (M + 1) + n = 0. So n = 3M + 6. (II) from (I), we know that f '(x) = 3mx2-6 (M + 1) x + 3M + 6 = 3M (x-1) [x - (1 + 2m)] when m < 0, there is 1 > 1 + 2m, when x becomes