# Chapter 4 - Section 4.5 - Summary of Curve Sketching - 4.5 Exercises - Page 322: 45

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.

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