`f(x) = (x^2) - x - ln(x)` (a) Find the intervals on which `f` is increasing or decreasing. (b) Find the local maximum and minimum values of `f`. (c) Find the intervals of concavity and the inflection points.
2 Answers | Add Yours
... and has no inflection points.
The function is defined for x>0 and is infinitely differentiable.
`f'(x) = 2x - 1 - 1/x.`
f'(x) = 0 is equivalent to `2x^2 - x - 1 = 0,` or x=1, x=-1/2. The second x is out of domain.
(a) for x in (0, 1) f'(x) < 0 and f is decreasing, for x in `(1, +oo)` f'(x)>0 and f is increasing.
(b) therefore x=1 is a local minimum, f(1) = 0.
(c) `f''(x) = 2 + 1/x^2` which is >0. So f is concave upward.
We’ve answered 318,915 questions. We can answer yours, too.Ask a question