If the function y = f (x) (f (x) is not always 0) and the image of y = - f (x) is symmetric about the origin, then what function is y = f (x)? Wrong, can you do it better

If the function y = f (x) (f (x) is not always 0) and the image of y = - f (x) is symmetric about the origin, then what function is y = f (x)? Wrong, can you do it better

Let P1 (a, b) be an arbitrary point on y = f (x), P2 (- A, - b) be a point on y = - f (x), because B = f (a), and - B = - f (- a), so f (a) = f (- a), that is, f (x) = f (- x), so y = f (x) is an even function

If the function f (x) = ax + 2 (a > 0 and a is not equal to 1) passes through a (2,6), then f (3) =?

The function f (x) = ax + 2 (a > 0 and a is not equal to 1)
f(2)=a*2+2=6
2a=4
a=2
f(x)=2x+2
f(3)=2*3+2=8

Let f (x) = the cube of X + the square of ax - a square x + 1, G (x) = the square of ax - 2x + 1, where the real number a is not equal to 0
Let f (x) = the cube of X + the square of ax - a square x + 1, G (x) = the square of ax - 2x + 1, where the real number a is not equal to 0. If f (x) is an increasing function in the interval (a, a + 2) of G (x), find the value range of A

f'(x)=3x^2+2ax-a^2=(x+a)(3x-a),g'(x)=2ax-2.
a> 0, f '(x) > 0 holds on (a, a + 2), G' (a) = 2A ^ 2-2 > = 0, a > = 1
a

10. Let f (x) = 2x cosx, {an} be an arithmetic sequence with tolerance of π / 8, f (A1) + F (A2) + +f（a5）=5π
10. Let f (x) = 2x cosx, {an} be an arithmetic sequence with tolerance of π / 8, f (A1) + F (A2) + +If f (A5) = 5 π, then [f (A3)] 2-a1a5=______ ．
A 0 B 1/16*π2 C 1/8*π2 D 13/16*π2
Choose D
∵ sequence {an} is an arithmetic sequence with a tolerance of π / 8,
And f (A1) + F (A2) + +f(a5)=5π
2a1-cosa1+2a2-cosa2+2a3-cosa3+2a4-cosa4+2a5-cosa5=5π
∴2(a1+a2+…… +a5)-(cosa1+cosa2+…… +cosa5)=5π
∴(cosa1+cosa2+…… +cosa5)=0
That is 2 (a1 + A2 +...) +a5)=2×5a3=5π,
a3=π/2,
a1=π/4
a5=3π/4
∴[f(a3)]²-a1a5
=(2a3-cosa3)²-a1a5
=(2*π/2-cosπ/2)²-π/4*3π/4
=π²-3π²/16
=13π²/16
My question is "cosa1 + cosa2 +" +Cosa5) = 0 "how did you get there?

F (x) = 2x cosx, so f (A1) = 2a1-cosa1f (A2) = 2a2-cosa2 +f(a5)=5π2(a1+a2+…… +a5)-(cosa1+cosa2+…… +Cosa5) = 5 π notice here because A5 = a1 + 4 * π / 8 = a1 + π / 2cosa1 + cosa2 + cosa3 + cosa4 + cosa5 = [cos (A3 -...)

Let f (x) = 2x cosx, {an} be an arithmetic sequence with tolerance of π / 8, f (A1) + F (A2) + If f (A5) = 5 π, then [f (A3)] ^ 2-a1 × A5 =?
May I ask why these formulas are finally changed to 10a3-cosa3 (1 + radical 2 + radical 2 + 2), because an is a series of numbers with π △ 8 as the equal difference, so cosa3 (1 + radical 2 + radical 2 + radical 2) does not contain π?

∵ sequence {an} is an arithmetic sequence with a tolerance of π / 8,
And f (A1) + F (A2) + +f(a5)=5π
2a1-cosa1+2a2-cosa2+2a3-cosa3+2a4-cosa4+2a5-cosa5=5π
∴2(a1+a2+…… +a5)-(cosa1+cosa2+…… +cosa5)=5π
∴(cosa1+cosa2+…… +cosa5)=0
That is 2 (a1 + A2 +...) +a5)=2×5a3=5π,
a3=π/2,
a1=π/4
a5=3π/4
∴[f(a3)]²-a1a5
=(2a3-cosa3)²-a1a5
=(2*π/2-cosπ/2)²-π/4*3π/4
=π²-3π²/16
=13π²/16

Given that the function f (x) is a monotone increasing function on R and is an odd function, the sequence {an} is an arithmetic sequence, A3 ＞ 0, then the value of F (A1) + F (A3) + F (A5) ()
A. Constant positive B. constant negative C. constant 0d. Can be positive or negative

∵ function f (x) is an odd function on R and an increasing function sequence. Take any x2 > x1, there is always f (x2) > F (x1), ∵ function f (x) is an odd function on R, ∵ function f (x) is an odd function on R and an increasing function, ∵ if x > 0, f (0) > 0, if x < 0, f (0) < 0

It is known that the domain of definition of odd function f (x) is (- 1,1) and monotonically decreasing, satisfying f (1-A) + the square of F (1-A)

Original formula = f (1-A) f (1-A ^ 2)

If we know the value range of [2xa + a] = 1, then what is the monotone function?

f(x)=x²-2x+3
=(x-1)²+2
The axis of symmetry is a straight line x = 1
So when a ≥ 1, there must be f (a + 1) > F (a), so it is a monotone increasing function
When a + 1 ≤ 1
When a ≤ 0, there must be f (a + 1)

Given that the domain of definition of function f (x) = √ (AX ^ 2-6ax + A + 8) is r, find the value range of real number a

ax^2-6ax+a+8≥0
a=0
ax^2-6ax+a+8=8≥0
a>0
ax^2-6ax+a+8=a(x^2-6x)+a+8=a(x-3)^2-9)+a+8=a(x-3)^2-8a+8
-8a+8≥0 1≥a
1≥a>0
1≥a≥0

If the function f (x) = log3a (x-1) defined in the interval (1,2) satisfies f (x) > 0, then the value range of a is___ ．

Because x ∈ (1,2), so X-1 ∈ (0,1), from F (x) > 0, we get 0 < 3A < 1, so 0 < a < 13, so the answer is: (0,13)