If f (x) satisfies 2F (x) - f (- x) = lgx, find f (x)

If f (x) satisfies 2F (x) - f (- x) = lgx, find f (x)

This topic is a mistake
2f(x)-f(-x)=lgx,
Then 2F (- x) - f (x) = LG (- x)
If lgx is significant, then x > 0
If LG (- x) is significant, - x > 0
There is a contradiction between the two

Let f (x) = the absolute value of lgx, a and B be real numbers satisfying f (a) = f (b) = 2F [(a + b) / 2], where 0

Let f (x) = | ㏒ 10 x |, a, B satisfy that f (a) = f (b) = 2F ((a + b) / 2) and 0 < a < B. prove: 3 < B < 2 + 2 under root sign
Can you solve it directly?
The image of F (x) should be well drawn
Obviously, a and B are on both sides of x = 1
Calculate... LGA = - LGB
a=1/b
Similarly, LGB = 2|lg (a + b) / 2|
There are two cases (a + b) / 21
And then there's a third factorization... It's easier
One answer is 3 + √ 13

Given the absolute value of F (x) = lgx, compare the values of F (1 / 4), f (1 / 3) and f (2)

When x1, f (x) = | lgx | = lgx
therefore
f(1/4)=lg4,
f(1/3)=lg3,
f(2)=lg2
Because f (4) > F (3) > F (2)
So f (1 / 4) > F (1 / 3) > F (2)