19: The domain is
D( f ) = (−1,0) ∪ (0,∞).
The derivative is
Critical points: The equation
f ′(x) = 0 cannot be solved analytically. Since
the denomintor is always positive in the domain, the sign of
f ′ is determined by its numerator
g(x) = x − (x + 1)⋅ln(x + 1).
We can guess that g = 0 for
x = 0.
However, we still cannot handle g analytically. Trick: We check on
how g goes. We find
g′(x) = −ln(x + 1), its signs tell us
that g is increasing on (−1,0) and decreasing on
(0,∞). This shows that
g(0) = 0 is a local maximum and also a global maximum
and so g is never positive for
x > −1.
We just learned that there are no critical points, so the intervals of
monotonicity will be
(−1,0) and
(0,∞), and also that
f ′ < 0 there. This should make the
monotonicity chart obvious.
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Answer