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Ask a math problem of senior one (more difficult)?
1、

The domain is R.

Make x1> x2

f(x 1)-f(x2)

=(2^x 1- 1)/(2^x 1+ 1)-(2^x2- 1)/(2^x2+ 1)

Denominator after general division = (2×1+1) (2× 2+1) > 0.

Molecule = (2x1-1) (2x2+1)-(2x2-1) (2x1+1)

=2^x 1*2^x2-2^x2+2^x 1- 1-2^x 1*2^x2-2^x2+2^x 1+ 1

=2*(2^x 1-2^x2)

x 1 & gt; x2

So 2 x 1-2 x2 > 0.

So the molecule is greater than zero.

So x1> F (x1) at x2 >: f (x2)

So its function is getting stronger and stronger.

2、

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

Multiply 2 x up and down, 2-x * 2 x = 1.

So f (-x) = (1-2x)/(1+2x) =-(2x-1)/(2x+1) =-f (x).

The domain is r, which is symmetrical about the origin.

So this is a strange function.