Increasing function on [0, positive infinity], then the range of values of real numbers A and B is

Solution: when a < 0, the function f(x)=a|x-b|+2 is a decreasing function in the interval [b, +∞] from a certain point in the direction of the positive real axis.

Therefore, only a > 0 can satisfy the meaning of the question, and the function f(x)=a|x-b|+2 is increasing function at [b, +∞], so only b≤0 can satisfy the meaning of the question.

Conclusion a > 0 and b≤0.

f(x)=[ 10 x- 10(-x)]/[ 10 x- 10(-x)]

It is proved that f(x) is a increasing function over a domain.

Prove that f(x) is multiplied by10 x up and down.

f(x)=( 10^2x- 1)/( 10^2x+ 1)=( 10^2x+ 1-2)/( 10^2x+ 1)

=( 10^2x+ 1)/( 10^2x+ 1)-2/( 10^2x+ 1)

= 1-2/( 10^2x+ 1)

Because 10 2x > 0, the denominator is not 0.

So the domain is R.

Make a>b

Then f (a)-f (b) =1-2/(102a+1)-1+2/(102b+1).

=2[( 10^2a+ 1)-( 10^2b+ 1)]/( 10^2a+ 1)( 10^2b+ 1)

Denominator is obviously greater than 0.

( 10^2a+ 1)-( 10^2b+ 1)= 10^2a- 10^2b

a & gtb，2a & gt2b

So102a-12b > 0.

so 2[( 10 2A+ 1)-( 10 2B+ 1)]/( 10 2A+ 1)( 10 2B+65438)。 0

Namely a>b stone

f(a)>f(b)

So f(x) is the increasing function in the domain.

The function f (x) = 2 x+2 (-x) a (constant a∈R) is known.

If a≤4, it is proved that f(x) is a increasing function on [1, +∞).

Proof: (Definition) Let 1 ≤ x 1

Then f (x1)-f (x2) = (2x1-2x2) (1-a/(2 (x1+x2)))

∫ 1≤x 1 & lt； x2

∴2≤x 1+x2

morever

2^(x 1+x2)≥4

If a & lt=4

Then 1-a/(2 (x 1+x2)) ≥ 0.

(2 x1-2x2) < 02 x is the increasing function on R.

∴f(x 1)-f(x2)=(2^x 1-2^x2)( 1-a/(2^(x 1+x2)))≤0

f(x 1)≤f(x2)

When a