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What is the coordinates of the intersection point between the image of the inverse function of F (x) = log3 (x + 3) and the Y axis?
y=log3 (x+3)
x+3=3＾y
x=3＾y-3
The inverse function is y = 3 ＾ x-3
When x = 0
y=1-3=-2
So the coordinates of the intersection point are (0, - 2)
The coordinates of the intersection point between the image of the inverse function of the function f (x) = log3 (x + 3) and the y-axis is______ .
Method 1: from the function f (x) = log3 (x + 3), the inverse function is y = 3x-3, let x = 0, y = - 2, that is, the coordinates of the intersection point between the image of the inverse function of the function f (x) = log3 (x + 3) and the Y axis is (0, - 2); method 2: from the known function f (x) = log3 (x + 3) image and X axis intersection point is (- 2, 0), because
The coordinates of the intersection point between the image of the inverse function of the function f (x) = log3 (x + 3) and the y-axis is______ .
Method 1: from the function f (x) = log3 (x + 3), the inverse function is y = 3x-3, let x = 0, y = - 2, that is, the coordinates of the intersection point between the image of the inverse function of the function f (x) = log3 (x + 3) and the Y axis is (0, - 2); method 2: from the known function f (x) = log3 (x + 3) image and X axis intersection point is (- 2, 0), because
Given that the function f (x) = (1-2x) / (1 + x), the image of function g (x) and the image of inverse function of function y = f (x + 1) are symmetric with respect to the straight line y = x, then G (2) =?
f(x)=(1-2x)/(1+x)
y=f(x+1)=(1-2x-2)/(2+x)=(-1-2x)/(2+x)
The image of function g (x) and the image of inverse function of function y = f (x + 1) are symmetric with respect to the straight line y = X,
Let (- 1-2x) / (2 + x) = 2
x=-5/4
That is g (2) = - 5 / 4
The inverse function image of function y = - (2x-1) / (x-3) is symmetric about points The inverse function image of the function y = - (2x-1) / (x-3) is symmetric about the point
The inverse function is y = 3-5 / (x + 2)
On (- 2,3) symmetry
If the function f (x) is in the inverse function and the tangent equation of the function f (x) image at the point (x, f (x)) is 2x-y + 1 = 0, then the image of the inverse function is at the point (f (x), x) In There is
That is to find the inverse function of 2x-y + 1 = 0
The tangent equation of the inverse function image at point (f (x), x) is as follows:
y=0.5x-0.5
Let f (x) and G (x) have inverse functions, and the images of F (x-1) and G inverse (X-2) are symmetric with respect to the straight line y = X. if G (5) = 2007, then f (4)
∵g(5)=2007
/ / g inverse (2007) = 5
/ / g inverse (X-2) over point (2009,5)
∵ the images of F (x-1) and G inverse (X-2) are symmetric about the line y=x
The point (52009) is on f (x-1)
That is, f (5-1) = f (4) = 2009
If the inverse function and the original function image are symmetric about the y = x axis, what are the requirements for the definition domain? Must it be r
There is no requirement that the inverse function and the original function image are axisymmetric with respect to y = X
If the definition domain and range of values are R, f (x + 2) is an odd function, FX has an inverse function, and the image of GX and FX is symmetric about y = x, then GX + G (- x) is equal to
G（x）+G（-x）=4
Given that the definition domain of the function y = f (2x-1) is an odd function of R, and the function y = g (x) is the inverse function of the function y = f (x), then G (a) + G (- a)=
0
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