find the domain of the function and where f is increase and decrease?

I am assuming this is:

f(x) = ln[(2 + x)/(2 - x)]

1)

This function will be defined when the expression under the log is positive i.e.:

(2 + x)/(2 - x) > 0

The critical values of this are x = -2 and x = 2. With sign diagrams, we get:

. . . . . . . . . . . .(-∞, -2). . . . . . . . . . . .(-2, 2). . . . . . . . . . . .(2, ∞)

Sign of 2 + x . . . . - . . . . . . . . . . . . . . . + . . . . . . . . . . . . . . +

Sign of 2 - x . . . . .+ . . . . . . . . . . . . . . .+ . . . . . . . . . . . . . . -

Sign of quotient . . - . . . . . . . . . . . . . . . + . . . . . . . . . . . . . . -

Therefore, ln[(2 + x) / (2 - x)] is defined only on the interval (-2, 2)

We can break this up to get:

f(x) = ln(2 + x) - ln(2 - x)

Then, we obtain:

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

f'(x) = (2 - x)/[(2 + x)(2 - x)] + (2 + x)/[(2 + x)(2 - x)]

f'(x) = 4/[(2 + x)(2 - x)]

This function decreases when f'(x) > 0 and decreases when f'(x) < 0

4/[(2 + x)(2 - x)] > 0

Since the numerator is always positive, we only need to consider the denominator.

(2 + x)(2 - x) > 0

Referring to our sign chart above, this increases on (-2, 2). We cannot consider anything else as they are outside of the domain.

I hope this helps!