Answer
(a) $y = 1$ is a horizontal asymptote.
(b) $f$ is decreasing on the interval $(-\infty,0)$
$f$ is increasing on the interval $(0,\infty)$
(c) The local minimum is $f(0) = -1$
(d) The graph is concave down on these intervals: $(-\infty,-\frac{2\sqrt{3}}{3})\cup (\frac{2\sqrt{3}}{3},\infty)$
The graph is concave up on this interval: $(-\frac{2\sqrt{3}}{3},\frac{2\sqrt{3}}{3})$
The points of inflection are $(-\frac{2\sqrt{3}}{3}, -\frac{1}{2})$ and $(\frac{2\sqrt{3}}{3}, -\frac{1}{2})$
(e) We can see a sketch of the graph below.
Work Step by Step
(a) $f(x) = \frac{x^2-4}{x^2+4}$
$f(x)$ is defined for all $x$ so there are no vertical asymptotes.
$\lim\limits_{x \to -\infty}f(x) = 1$
$\lim\limits_{x \to \infty}f(x) = 1$
$y = 1$ is a horizontal asymptote.
(b) We can find the points where $f'(x) = 0$:
$f'(x) = \frac{(2x)(x^2+4)- (x^2-4)(2x)}{(x^2+4)^2}$
$f'(x) = \frac{2x^3+8x-2x^3+8x}{(x^2+4)^2}$
$f'(x) = \frac{16x}{(x^2+4)^2} = 0$
$16x = 0$
$x = 0$
When $x \lt 0~~$ then $f'(x) \lt 0$
$f$ is decreasing on the interval $(-\infty,0)$
When $x \gt 0$ then $f'(x) \gt 0$
$f$ is increasing on the interval $(0,\infty)$
(c) $f(0) = \frac{0^2-4}{0^2+4} = -1$
The local minimum is $f(0) = -1$
There is no local maximum.
(d) We can find the points where $f''(x) = 0$:
$f''(x) = \frac{(16)(x^2+4)^2-(16x)(2)(x^2+4)(2x)}{(x^2+4)^4}$
$f''(x) = \frac{(16)(x^4+8x^2+16)-(64x^2)(x^2+4)}{(x^2+4)^4}$
$f''(x) = \frac{16x^4+128x^2+256-64x^4-256x^2}{(x^2+4)^4}$
$f''(x) = \frac{-48x^4-128x^2+256}{(x^2+4)^4}$
$f''(x) = \frac{(-48x^2+64)(x^2+4)}{(x^2+4)^4}$
$f''(x) = \frac{-48x^2+64}{(x^2+4)^3} = 0$
$48x^2= 64$
$x^2 = \frac{4}{3}$
$x = \pm \frac{2\sqrt{3}}{3}$
The graph is concave down when $f''(x) \lt 0$
The graph is concave down on these intervals: $(-\infty,-\frac{2\sqrt{3}}{3})\cup (\frac{2\sqrt{3}}{3},\infty)$
The graph is concave up when $f''(x) \gt 0$
The graph is concave up on this interval: $(-\frac{2\sqrt{3}}{3},\frac{2\sqrt{3}}{3})$
$f(-\frac{2\sqrt{3}}{3}) = \frac{(-\frac{2\sqrt{3}}{3})^2-4}{(-\frac{2\sqrt{3}}{3})^2+4} = -\frac{1}{2}$
$f(\frac{2\sqrt{3}}{3}) = \frac{(\frac{2\sqrt{3}}{3})^2-4}{(\frac{2\sqrt{3}}{3})^2+4} = -\frac{1}{2}$
The points of inflection are $(-\frac{2\sqrt{3}}{3}, -\frac{1}{2})$ and $(\frac{2\sqrt{3}}{3}, -\frac{1}{2})$
(e) We can see a sketch of the graph below.