Let the function f (x) be a function with t as the period, prove that f (AX + b) (A and B are positive numbers) is also a periodic function, and find its period

f(x)=f(x+t)
f(ax+b)=f(ax+b+t)=f[a(x+t/a)+b]
So it's a periodic function with period = t / | a |

It is proved that if f (x) is a periodic function with L as the period, then f (AX + b) (a, B are constants, and a > 0) is a periodic function with L / A as the period

If the period of F (x) is I, then according to the definition, f (x + Ki) = f (x), that is:
If y = x + Ki, then f (y) = f (x);
For the function: G (x) = f (AX + b),
When y = x + K (I / a), ay + B = a [x + K (I / a)] + B = ax + B + ki
g(y)=f(ay+b)=f(ax+b+kI)
According to the periodic property of F (x), f (AX + B + Ki) = f (AX + b) = g (x)
So there are:
When y = x + K (I / a), G (y) = g (x)
That is, G (x + K (I / a)) = g (x)
So g (x) = f (AX + b) is a periodic function with I / A as the period

Let f (x) be a function with t as the period and a be any positive real number. It is proved that f (AX) is a function with T / A as the period

F (x) is a function of period T
Then f (x + T) = f (x)
So f (AX + T) = f (AX)
And f (AX + T) = f [a (x + T / a)] = f (AX)
That is, in F (AX), any x increases the T / a unit, and the function value repeats
‡ f (AX) is a periodic function with a period of T / A

High school mathematics - periodic function: please prove that 'if f (x) satisfies f (x + T) = 1 / F (x), then f (x) is a periodic function with a period of 2T' The more detailed, the better! Please ~ thank you! ^^

For any x, f (x + 2t) = 1 / F (x + T) = 1 / [1 / F (x)] = f (x), so f (x) is a periodic function with a period of 2T

For function f (x), if there is x0 < R and f (x0) = x0 holds, x0 is called the fixed point of F (x). It is known that function f (x) = ax ∨ 2 + (B + 1) x + (B-1) (a ≠ 0) 1) When a = 1 and B = 2, find the fixed point of function f (x) 2） If for any real number B, the function always has two different fixed points, find the value range of A 3) Under the condition of 2, if the abscissa of two points a and B on the image of y = f (x) is the fixed point of function f (x), and the two points a and B are symmetrical about the square + 1 of y = KX + 1 / 2a, find the minimum value of B

Let y = f (x)
The fixed point F (x0) = x0 is actually the intersection of the function y = f (x) image and y = X
1. When a = 1, B = 2,
y=F(X)=aX^2 +(b+1)X+(b-1)=x^2+3x+1
y=x
The solution is x = - 1, y = - 1
2.f(x)=ax^2 +(b+1)x+(b-1)=x
ax^2 +bx+(b-1)=0
According to the problem, this equation has two different real number solutions, namely Δ＞ 0
b^2-4*a*(b-1)＞0
b^2-4ab+4a＞0
If the above formula is constant, that is, the minimum value of the quadratic function B ^ 2-4ab + 4A about B is greater than 0
That is, (4 * 4a-4a * 4a) / 4 = 4a-4a ^ 2 > 0
Solving inequality a < 0 or a > 4
three

For function f (x), if there is x0 ∈ r so that f (x0) = x0, x0 is called the fixed point of F (x). Given that function f (x) = AX2 + (B + 1) x + (B-1) (a ≠ 0), if function f (x) always has two different fixed points for any real number B, the value range of a is ___

F (x) = AX2 + (B + 1) x + B-1 = x has two unequal real roots,
‡ AX2 + BX + B-1 = 0 has two unequal real roots,
The discriminant is always true when it is greater than 0, that is, b2-4a (B-1) > 0
∴△=（-4a）2-4 × 4a＜0
∴0＜a＜1，
The value range of a is 0 ＜ a ＜ 1
So the answer is 0 < a < 1

Let the function f (x) be a function with t as the period, prove that f (AX + b) (A and B are positive numbers) is also a periodic function, and find its period