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If the function y = (x) has an inverse function y = F-1 (x) and the image of function y = 2x-f (x) passes (2,1), then the function y = F-1 (x)- 2X image must pass through the point___
By substituting (2,1) into y = 2x-f (x), f (2) = 3, that is, F-1 (3) = 2
Substituting x = 3 into the function y = F-1 (x) - 2x
y=f-1(3)-6=2-6=4
So the function must pass through the point (3,4)
It is known that the function y = f (x) has an inverse function y = f ^ - 1 (x). If the image of function y = f (x + 1) passes through point (3,1), then the image of function y = F-1 (x) must pass through the point How does this go around?
The image of y = f (x + 1) passes (3,1)
That is, f (4) = 1,
So the image of y = f (x) passes through (4,1)
So the inverse function y = f ^ - 1 (x) passes (1,4)
Let y = f (x) have an inverse function y = F-1 (x), and the function y = F-1 (x) - x, the image crossing point (2,1) of the function y = f (x) exists, Then the image of the function y = x-f (x) must pass through the point?
Image crossing point (2,1) of y = F-1 (x) - x
That is, 1 = F-1 (2) - 2
f-1(2)=3
Therefore, f (3) = 2
Let x = 3 in y = x-f (x)
Y = 3-F (3) = 1
So the image of y = x-f (x) must pass through the point (3,1)
28.2 if the inverse function f - 1 (x) of the function y = f (x) = (1-2x) / (3 + x) (x ∈ R and X ≠ - 3), then y = the image (b) of F (x) If the inverse function f - 1 (x) of the function y = f (x) = (1-2x) / (3 + x) (x ∈ R and X ≠ - 3), then y = the image (b) of F (x) (A) Symmetry about point (2,3) (b) symmetry about point (- 2, - 3) (C) On the symmetry of a straight line (y = 3)
Let F-1 (x) = y
Then the positions of X and y are interchanged
x=(1-2y)/(3+y)=[-2(3+y)+7]/(3+y)
=7/(3+y)-2
x+2=7/(y+3)
y+3=7/(x+2)
y=7/(x+2)-3
According to the function image, the center of symmetry is a point (- 2, - 3)
Let the inverse function of the function y = f (x) be y = F-1 (x) and the image crossing point (1) of y = f (2x-1) Then y = F-1 (x)______ .
∵ y = f (2x-1)
2，1)，
ν y = f (x) image crossing point (0, 1),
According to the images of two functions which are inverse functions of each other, they are symmetric with respect to the straight line y = X,
The image crossing point (1,0) of y = F-1 (x) is obtained
Therefore, fill in: (1, 0)
It is known that the inverse function of function y = f (x) is y = f ^ - 1 (x), and the inverse function of function y = f (2x-1) + 1 is F-1 (x-1) / 2 + 1 / 2. Why
y=f(2x-1)+1
y-1=f(2x-1)
After inverse function
F ^ - 1 (x-1) = 2y-1 (x, y is only a form, in this case, if it still means f ^ - 1 (Y-1) = 2x-1, it is easy to confuse)
y=[f^-1(x-1)+1]/2
(2004 Wuhan simulation) given that the image of function y = F-1 (x) is over (1,0), then y = f (1) The image of the inverse function of 2x − 1) must pass through the point () A. （1，2） B. （2，1） C. （0，2） D. （2，0）
Resolution: ∵ the image of function y = F-1 (x) passes (1,0),
∴f1（1）=0，⇒f（0）=1
Then the graph of function f (x) passes through (0,1) point,
The function y = f (1
The image of 2x − 1) must pass through (2,1) points,
Then y = f (1)
The image of the inverse function of 2x − 1) must pass the point (1,2)
Therefore, a
Let y = f (x) and G (x) have inverse functions, and the images of F (x-1) and G ^ - 1 (X-2) functions are symmetric with respect to the straight line y = X. if G (5) = 2006, calculate the value of F (4)
g(5)=2006
g^-1(2006)=5
g^-1(2008-2)=5
Image crossing point of function G ^ - 1 (X-2) (2008,5)
The image crossing point of function f (x-1) (52008)
f(5-1)=2008
That is, f (4) = 2008
Let f (x) = a ^ x (where a > 0 and a ≠ 1) (1), find the analytic expression of the inverse function φ (x) of the function y = f (x)
The inverse of the exponential function is the logarithmic function,
Therefore, φ (x) = loga (x)
It is known that the inverse function of the function f (x) = log2x (x > 0) is F-1 (x), and there is F-1 (a) · F-1 (b) = 2. If a, b > 0, then the minimum value of 1 / A + 4 / B is_____ I figured it was 8, but the answer was 9. It's in several books
It is known that the inverse function of the function f (x) = log2x (x > 0) is F-1 (x), and there is F-1 (a) · F-1 (b) = 2. If a, b > 0, then the minimum value of 1 / A + 4 / B is_____
The inverse function of the function f (x) = log2x (x > 0) is F-1 (x) = 2 ^ X
F-1 (a) · F-1 (b) = 2
2^(a+b)=2^1
a+b=1
1/a+4/b=(a+b)/a+4(a+b)/b
=1+b/a+4a/b+4
=5+b/a+4a/b>=5+4=9
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