Catastrophe!! The master is coming!! f a x f a x 2b, y f x , analytic expression about x 2 symmetry

Necessity) set point p(x,y) is any point on the image of y = f (x), point p( x ,y) symmetry point p'(2a x, 2b y) with respect to the point a (a ,b) is also on the y = f (x) image, 2b y = f (2a x) i.e. y + f (2a x) = 2b so f (x) + f (2a x) = 2b, necessity is proved.

Sufficiency) set point p(x0,y0) is y = f (x) any point on the image, then y0 = f (x0).

f (x) + f (2a x) = 2b f (x0) + f (2a x0) = 2b, i.e. 2b y0 = f (2a x0).

Therefore, the point p'(2a x0,2b y0) is also on the y = f (x) image, and the point p and the point p' are symmetrical with respect to the point a (a, b), and the sufficiency is proved.

with respect to x=2 symmetry, then the new analytic formula satisfies :

f(4-(a+x))+f(4-(a-x))=2b => f(4-a-x)+f(4-a+x)=2b

That is, for each point on the image, the ordinate is unchanged, and the abscissa is taken as the symmetry point about x=2. (The symmetry point of x is 4-x).

f(4-a+x)+f(4-a-x)=2b

There is the original formula f(x) about (symmetry.

f(x) is symmetrical with respect to x=2 then f''(x) is symmetrical with respect to (4-a,b), so.

f(4-a+x)+f(4-a-x)=2b

Let x=1-t

Then -x=t-1

2-x=t+1

So there is f(1+t)=f(1-t).

i.e. f(1+x) = f(1-x).

It means that the values of the +x,-x function on both sides of x=1 are equal, so the values are symmetrical with respect to x=1.

x=(a-b)/2

Let a-x1=x2-b

Get x1+x2=a+b

Hence symmetry with respect to x=(a-b) 2.

I don't know about the rest, I'm sorry.

The image of the function f(x) and y=a x is symmetrical with respect to y=x, f(x)=log(a is the base), x

g(x)=[log(a) x]*

Let t=log(a is the base) x

g(x)=t^2+t

Axis of symmetry 1 2-[log(a is the base)2].

The opening is upward. If g(x) is an increasing function on [1 2,2], then the axis of symmetry < = 1 2

1 2 - [log(a is the base)2]< = 1 2

log(a is the base)2]>=0

Get a>1