The relationship between two function images which are reciprocal functions

On y = x symmetry

The odd function y = f (x) on R has an inverse function y = F-1 (x). If y = f (x + 1) and y = F-1 (x + 2) are inverse functions, then f (2009)=
For detailed explanation, the process should not be omitted

Y = f (x + 1) and y = F-1 (x + 2) are inverse functions,
∴f(x+1)=f(x)-2,
F (x) is an odd function on R,
∴f(0)=0,f(n)=-2n,n∈N,
∴f（2009）=-2*2009=-4018.

The image of odd function y = f (x) (x ∈ R) must pass through point ()
A. （a，f（-a））B. （-a，f（a））C. （-a，-f（a））D. (a，f(1a))

According to the function y = f (x) is an odd function, we can get f (- x) = - f (x), so f (- a) = - f (a), so the image of the function passes through the point (- A, f (- a)), that is, the image of the function passes through the point (- A, - f (a)), so we choose C

Given that the function y = f (x) is an odd function, when x ≥ 0, f (x) = 3x-1, let the inverse function of F (x) be y = g (x), then G (- 8)=______

Method 1: when X & lt; 0, - X & gt; 0, f (- x) = 3-x-1. And ∵ f (x) is an odd function, ∵ f (- x) = - f (x), that is - f (x) = 3-x-1. ∵ f (x) = 1-3-x. ∵ f (x) = 3x − 11 − 3 − XX ≥ 0x & lt; 0. ∵ F-1 (x) = log3 (x + 1) & nbsp; & nbsp; & nbsp; & nbsp

Let the function y = f (x) be continuous and monotone on [a, b], and prove that its inverse function is also continuous and monotone on the corresponding interval

Let e be any point between ab
a

Given the function f (x) = ax & # 178; + BX + C (a ≠ 0), the image passes through a (0,1) and (- 1,0), tangent B & # 178; - 4A ≤ 0
① Find the analytic expression of F (x);
② Under the condition of (1), when x ∈ [- 2,2], G (x) = f (x) - KX is a monotone function

① Because f (x) passes through (0,1) (- 1,0), the coordinates of two points are substituted into f (x)
c = 1 ; a - b + c = 0.
B ^ 2 - 4A = 2 or 1-k / 2

Given the quadratic function y = ax & # 178; + BX + C, ad, where a > 0, B & # 178; - 4A & # 178; C & # 178; = 0, its image has only one intersection with the x-axis, the intersection is a, the intersection with the y-axis is b, and ab = 2
(1) Find the analytic expression of quadratic function;
(2) When B

1) There is only one intersection point between Y and X axis;
This point must be the extreme point of quadratic function y, that is, y '= 2aX + B = 0
X = - B / 2a, that is, the intersection point with X axis is [- B / 2a, 0]
The point of intersection with Y-axis is obviously [0, C]
∵AB=2
That is, B & # 178; + 4A & # 178; C & # 178; = 16A & # 178; (1)
∵ y = ax & # 178; + BX + C = 0 has only one root;
∴b²-4ac=0 .(2)
∵b²-4a²c²=0.(3)
According to (2) (3), AC = 1
By substituting (4) into (2), we can get that B = ± 2
Substituting (3) into (1), we can get 8A & # 178; C & # 178; = 16A & # 178;
That is a & # 178; = 1 / 2
∵a>0
∴a=√2/2
∴c=√2
The analytic expression of quadratic function is y = √ 2 / 2x & # 178; ± 2x + √ 2
2) According to the meaning B

Let f (x) = x & # 179; ax & # 178; bxc, we know that f (x) is an odd function and its image passes through point (2,0)
Let g (x) = f (x + 1) / x + 1. If G (a) < g (2) holds, the value range of a is obtained

Is the expression of the function determined like this or F (x) = x ^ (3a) x ^ (2b) x ^ (c)?

In the following functions, the one passing through the origin is a.y = 5x + 1, b.y = - 5x-1, C.Y = - X / 5

c.y=-x/5

As shown in the figure, the line passing through the origin o intersects the image of the function y = 2 ^ x at points a and B. the vertical lines passing through a and B respectively intersect the image of the function y = 4 ^ x at points c and D
Verification: O, C, C are collinear

Let a point (x1,2 ^ x1), B point (x2,2 ^ x2) make the straight line y = KX (the straight line should have two intersections with y = 2 ^ x, which needs to meet the following conditions: k > e), then: k = 2 ^ X1 / X1 = 2 ^ x2 / X2, then: C point (x, y), then: y = 2 ^ x1, that is: 4 ^ x = 2 ^ x1, that is: 2 ^ X1 (2x) = 2 ^ x1, that is: x = X1 / 2, then: B = 2 ^ X2, that is: 4 ^ a = 2 ^ X2, that is

The relationship between two function images which are reciprocal functions