Answer
A. The domain is $(-\infty,0) \cup (0, \infty)$
B. There is no y-intercept.
There are no x-intercepts.
C. The function is not an odd function or an even function.
D. $y = 1$ is a horizontal asymptote.
$x = 0$ is a vertical asymptote.
E. The function is decreasing on the intervals $(-\infty, -2)\cup (0,\infty)$
The function is increasing on the interval $(-2,0)$
F. The local minimum is $(-2, \frac{3}{4})$
G. The graph is concave up on the interval $(-3,0)\cup (0,\infty)$
The graph is concave down on the interval $(-\infty, -3)$
The point of inflection is $(-3, \frac{7}{9})$
H. We can see a sketch of the curve below.
Work Step by Step
$y = 1+\frac{1}{x}+\frac{1}{x^2}$
A. The function is defined for all real numbers except $x = 0$
The domain is $(-\infty,0) \cup (0, \infty)$
B. There is no y-intercept because $x \neq 0$
When $y = 0$:
$1+\frac{1}{x}+\frac{1}{x^2} = 0$
$\frac{x^2+x+1}{x^2} = 0$
There are no values of $x$ such that $y = 0$
There are no x-intercepts.
C. The function is not an odd function or an even function.
D. $\lim\limits_{x \to -\infty} (1+\frac{1}{x}+\frac{1}{x^2}) = 1$
$\lim\limits_{x \to \infty} (1+\frac{1}{x}+\frac{1}{x^2}) = 1$
$y = 1$ is a horizontal asymptote.
$\lim\limits_{x \to 0^-} (1+\frac{1}{x}+\frac{1}{x^2}) = \infty$
$\lim\limits_{x \to 0^+} (1+\frac{1}{x}+\frac{1}{x^2}) = \infty$
$x = 0$ is a vertical asymptote.
E. We can find the values of $x$ such that $y' = 0$:
$y' = -\frac{1}{x^2}-\frac{2}{x^3} = 0$
$-\frac{1}{x^2} = \frac{2}{x^3}$
$x = -2$
When $x \lt -2$ or $x \gt 0$, then $y' \lt 0$
The function is decreasing on the intervals $(-\infty, -2)\cup (0,\infty)$
When $-2 \lt x \lt 0$, then $y' \gt 0$
The function is increasing on the interval $(-2,0)$
F. When $x = -2$, then $y = 1+\frac{1}{(-2)}+\frac{1}{(-2)^2} = \frac{3}{4}$
The local minimum is $(-2, \frac{3}{4})$
G. We can find the values of $x$ such that $y'' = 0$:
$y'' = \frac{2}{x^3}+\frac{6}{x^4} = 0$
$2x+6= 0$
$x = -3$
When $-3 \lt x \lt 0~~$ or $x \gt 0$, then $y'' \gt 0$
The graph is concave up on the interval $(-3,0)\cup (0,\infty)$
When $x \lt -3$, then $y'' \lt 0$
The graph is concave down on the interval $(-\infty, -3)$
When $x = -3$, then $y = 1+\frac{1}{(-3)}+\frac{1}{(-3)^2} = \frac{7}{9}$
The point of inflection is $(-3, \frac{7}{9})$
H. We can see a sketch of the curve below.
