Answer
(a) $f$ is decreasing on the intervals $(-\infty, -1)\cup(1,\infty)$
$f$ is increasing on the interval $(-1,1)$
(b) The local maximum is $f(1) = \frac{1}{2}$
The local minimum is $f(-1) = -\frac{1}{2}$
(c) The graph is concave down on these intervals:
$(-\infty, -\sqrt{3})\cup (0, \sqrt{3})$
The graph is concave up on these intervals:
$(-\sqrt{3}, 0)\cup (\sqrt{3}, \infty)$
The points of inflection are $(-\sqrt{3},-\frac{\sqrt{3}}{4}), (0,0),$ and $(\sqrt{3},\frac{\sqrt{3}}{4})$
Work Step by Step
(a) $f(x) = \frac{x}{x^2+1}$
We can find the points where $f'(x) = 0$:
$f'(x) = \frac{(x^2+1)-(x)(2x)}{(x^2+1)^2}$
$f'(x) = \frac{1-x^2}{(x^2+1)^2} = 0$
$1-x^2 = 0$
$x^2 = 1$
$x = \pm 1$
When $x \lt -1$ and $x \gt 1,$ then $f'(x) \lt 0$
$f$ is decreasing on the intervals $(-\infty, -1)\cup(1,\infty)$
When $-1 \lt x \lt 1,$ then $f'(x) \gt 0$
$f$ is increasing on the interval $(-1,1)$
(b) $f(-1) = \frac{(-1)}{(-1)^2+1} = -\frac{1}{2}$
$f(1) = \frac{(1)}{(1)^2+1} = \frac{1}{2}$
The local maximum is $f(1) = \frac{1}{2}$
The local minimum is $f(-1) = -\frac{1}{2}$
(c) We can find the points where $f''(x) = 0$:
$f''(x) = \frac{(-2x)(x^2+1)^2-(1-x^2)(2)(x^2+1)(2x)}{(x^2+1)^4}$
$f''(x) = \frac{(-2x)(x^4+2x^2+1)-(2-2x^2)(2x^3+2x)}{(x^2+1)^4}$
$f''(x) = \frac{-2x^5-4x^3-2x-4x^3-4x+4x^5+4x^3}{(x^2+1)^4}$
$f''(x) = \frac{2x^5-4x^3-6x}{(x^2+1)^4} = 0$
$2x^5-4x^3-6x = 0$
$2x~(x^4-2x^2-3) = 0$
$2x~(x^2-3)(x^2+1) = 0$
$x = 0~~$ or $~~x = \pm \sqrt{3}$
The graph is concave down when $f''(x) \lt 0$
The graph is concave down on these intervals:
$(-\infty, -\sqrt{3})\cup (0, \sqrt{3})$
The graph is concave up when $f''(x) \gt 0$
The graph is concave up on these intervals:
$(-\sqrt{3}, 0)\cup (\sqrt{3}, \infty)$
$f(-\sqrt{3}) = \frac{-\sqrt{3}}{(-\sqrt{3})^2+1} = -\frac{\sqrt{3}}{4}$
$f(0) = \frac{0}{(0)^2+1} = 0$
$f(\sqrt{3}) = \frac{\sqrt{3}}{(\sqrt{3})^2+1} = \frac{\sqrt{3}}{4}$
The points of inflection are $(-\sqrt{3},-\frac{\sqrt{3}}{4}), (0,0),$ and $(\sqrt{3},\frac{\sqrt{3}}{4})$
