Answer
A. The domain is $(0, \infty)$
B. There is no y-intercept.
There are no x-intercepts.
C. The function is not an even function or an odd function.
D. $\lim\limits_{x \to 0^+} (\frac{1}{x}+ln~x) = \infty$
$x = 0$ is a vertical asymptote.
$\lim\limits_{x \to \infty} (\frac{1}{x}+ln~x) =\infty$
E. The function is decreasing on the interval $(0, 1)$
The function is increasing on the interval $(1, \infty)$
F. $(1,1)$ is a local minimum.
G. The graph is concave up on the interval $(0, 2)$
The graph is concave down on the interval $(2, \infty)$
The point of inflection is $(2, 1.19)$
H. We can see a sketch of the curve below.
Work Step by Step
$y = \frac{1}{x}+ln~x$
A. The function is defined for all real numbers such that $x \gt 0$
The domain is $(0, \infty)$
B. Since $x \neq 0,$ there is no y-intercept.
When $y = 0$:
$\frac{1}{x}+ln~x = 0$
$ln~x = -\frac{1}{x}$
There are no values of $x$ such that $~~ln~x= -\frac{1}{x}$
There are no x-intercepts.
C. The function is not an even function or an odd function.
D. $\lim\limits_{x \to 0^+} (\frac{1}{x}+ln~x) = \lim\limits_{x \to 0^+} \frac{1+x~ln~x}{x} = \infty$
$x = 0$ is a vertical asymptote.
$\lim\limits_{x \to \infty} (\frac{1}{x}+ln~x) =\infty$
There is no horizontal asymptote.
E. We can find the values of $x$ such that $y' = 0$:
$y' = -\frac{1}{x^2}+\frac{1}{x} = \frac{x-1}{x^2} = 0$
$x-1 = 0$
$x = 1$
When $0 \lt x \lt 1$, then $y' \lt 0$
The function is decreasing on the interval $(0, 1)$
When $x \gt 1$, then $y' \gt 0$
The function is increasing on the interval $(1, \infty)$
F. When $x = 1$, then $y = \frac{1}{1}+ln~1 = 1+0 = 1$
$(1,1)$ is a local minimum.
G. We can find the values of $x$ such that $y'' = 0$:
$y'' = \frac{x^2-(x-1)(2x)}{x^4}$
$y'' = \frac{x-(x-1)(2)}{x^3}$
$y'' = \frac{-x+2}{x^3} = 0$
$-x+2 = 0$
$x = 2$
When $0 \lt x \lt 2$, then $y'' \gt 0$
The graph is concave up on the interval $(0, 2)$
When $x \gt 2$, then $y'' \lt 0$
The graph is concave down on the interval $(2, \infty)$
When $x=2$, then $y = \frac{1}{2}+ln~2 = 1.19$
The point of inflection is $(2, 1.19)$
H. We can see a sketch of the curve below.