Problem description:

Are the function image and the inverse function image necessarily symmetrical about y=X? What is the reason?

There are also: y= (116) x and its inverse function "y = logarithm of x with116 as the base".

(1/2, 1/4) and (1/4, 1/2), why is the intersection not on the straight line with y=x? There seems to be a theorem that "the intersections of linearly symmetric graphs are all on the axis of symmetry", isn't there?

Analysis:

F~ 1(x) is always used as the inverse function of f(x).

The conclusion of 1 is correct, assuming that any point (a, b) on f(x) has f (a) = b.

Therefore, f~ 1(b)=a means that f~ 1(x) must pass (b, a) and (a, b, a) is symmetrical about the straight line y=x, that is, the abscissa and the ordinate are interchanged, so the image is symmetrical about the straight line y = X.

2 (1) This function and its inverse function are drawn as follows:

img 699 . photo . 163/jypqzone/* * * * * * * * * * * * * * * *

The red line represents the image of y = (116) x, and the blue line represents the image of y = log-116-x.

It seems to be tangent at a certain point, and the intersection point should be y = X.

But after verification, it is found that (1/2, 1/4) and (1/4, 1/2) are indeed the intersection of two functions.

However, they must also have an intersection point on y=x, which is proved as follows:

Intuitively, the line segment connecting (1/2, 1/4) and (1/4, 1/2) passes through the straight line y = x.

These two functions are continuous. If f(x) and the straight line y=x do not intersect, that is, f(x) is only on one side of y=x, then y=f~ 1(x) is on the other side of y=x, so they cannot intersect again.

Therefore, if they intersect in addition to y=x, there is at least one intersection P(x0, y0).

Such that x0=y0

Specific to this problem, it can be proved that y = (116) x and y=x have intersections in the interval (1/4, 1/2) as follows:

Let f (x) = (116) x-x make f(x)=0, then we can get the intersection of y = (116) x and y = X.

Their intersection must also be on y = log-116-x.

And f(x) is continuous in the interval (1/4, 1/2).

f( 1/4)= 1/4 & gt； 0 f( 1/2)=- 1/4 & lt； 0f( 1/4)* f( 1/2)& lt； 0

Therefore, f(x) must intersect the X axis at least 1 point in the interval (1/4, 1/2).

That is to say, at least one point P(x 1, y 1) makes f (x) = (116) x and f~ 1(x) intersect this point.

(2) Summary of "the intersection point is not on the straight line of y=x"

I come to the conclusion that if f(x) is defined, continuous and monotonically increasing in the interval I, then the intersection of F (x) and its inverse function must be on the straight line Y = X..

Proof: Suppose f(x) passes through point P(x0, y0) in its inverse function (x0 is not equal to y0).

That is, f(x0)=y0 f~ 1(x0)=y0 (f~ 1(x) is the inverse function of f(x) for the time being).

So f(y0)=x0 f~ 1(y0)=x0.

That is, both f(x) and f~ 1(x) pass through (x0, y0) (y0, x0).

Because x0 is not equal to y0, x0 is set.

Because f(x) passes through (x0, y0) (y0, x0).

It is found that the dependent variable decreases with the increase of independent variables, and it is monotonous in the interval.

Therefore, f(x) is a decreasing function in the interval.

What about you?

This contradicts that the topic condition f(x) is the increasing function in the interval.

Therefore, if there is an intersection x0 that is not equal to y0, that is, the intersection must be on the straight line y = X, it is wrong.

For the subtraction function in the interval, it cannot be proved and a counterexample is given, such as this:

img 699 . photo . 163/jypqzone/* * * * * * * * * * * * * * * *

But I have a guess about this problem: let the function f(x) be defined in the interval I and have continuous first and second derivatives in I, and there is no inflection point in I, that is, there is no T belonging to I, so that f "(t) = 0 or f(x) is always convex or concave in the interval.

Then the intersection of f(x) and its inverse function must be on the straight line y = X.

This is just a guess, I can't prove it. It's just that drawing looks right.

(3) The explanation of "the intersections of linear symmetric graphs are all on the axis of symmetry"

Personally, I think this is a wrong conclusion. For example, the subtraction function above is a counterexample.

(4) A "special case" explanation of "given that f(x) is defined on the interval I and continuously and monotonically increases, then the intersection of F (x) and its inverse function must be on the straight line y=x"

Someone gave an example: y=- 1/x is the increasing function in the domain.

Its inverse function is that it intersects itself at countless points that are not on y = X.

I explain it this way: don't forget that there is a definition of continuity in "I" X=0 is a discontinuous point of this function. If it is not defined at this point, it is not a "continuous function defined in I"

(5) Add points: the function and its inverse function do not necessarily have intersections. As long as f(x) and f~ 1(x) are on both sides of y=x, this need not be explained too much.

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