I feel like giving my take on this since it's an interesting question. Starting with
and
both being odd:
^{n-1}(x-b)^{m-1}((m+n)x - am - bn))
If
and
are both odd, then
^{m-1})
and
^{n-1})
are both raised to even powers and are therefore positive, except where they are zero, which is when
or when
For the linear factor, you can simply solve the inequality:
x - am - bn>0)
x >am + bn)
So the derivative is positive provided
,
and
For the case where
is instead even, the factor
is still positive except where
For the remaining two factors, they either need to be both positive, or both negative, for their product to be positive.
If they are both positive, we require that
(since it is raised to an odd power,
) i.e.
, and also that
, from solving the same inequality as before. So we require that
is greater than the larger of
and
Using the same method, if they are both negative, then
and
, so
must be less than the smaller of the two.
So overall we require that
\cup(\mathrm{max}\left\{{a,\frac{am+bn}{m+n}\right\},\infty)\right)\setminus \left\{b\right\})
if
is even and
is odd
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