39We first identify increasing decreasing sections of the domain.
f'(x) = xx(logx + 1)xx + 1
Put f'(x) = 0.
Either xx = 0 or logx + 1 = 0
As xx ≠0,
logx = -1
=> x = 1/e
Thus the function's derivative changes sign across x = 1/e.
We know that
2 < e < 3
=> 1/2 > 1/e > 1/3
So 1/e lies in (0, 1).
We dissect the domain into two pieces,
x → (0, 1/e] U (1/e, 1) where in the right part f(x) increases and in the left part f(x) decreases, which you can confirm by the wavy curve method.
=> f(x) → [ log[(1e)1e + 1], log2 ) U ( log[(1e)1e + 1], log2)
Which is actually
f(x) → [ log[(1e)1e + 1], log2)
Note that (0, 1/e] is the part where f(x) is decreasing, so when we apply 0 and 1/e as inputs to the function, we switch places with the outputs because a smaller input corresponds to a larger output(decreasing function property).