Chapter 4 - Section 4.5 - Summary of Curve Sketching - 4.5 Exercises - Page 322: 52

A. The domain is $(0, \infty)$ B. There is no y-intercept. The x-intercept is $1$ C. The function is not an even function or an odd function. D. $\lim\limits_{x \to 0^+} \frac{ln~x}{x^2} = -\infty$ $x = 0$ is a vertical asymptote. $\lim\limits_{x \to \infty} \frac{ln~x}{x^2} =0$ $y = 0$ is a horizontal asymptote. E. The function is increasing on the interval $(0, 1.65)$ The function is decreasing on the interval $(1.65, \infty)$ F. $(1.65,0.184)$ is a local maximum. G. The graph is concave down on the interval $(0, 2.3)$ The graph is concave up on the interval $(2.3, \infty)$ The point of inflection is $(2.3, 0.157)$ H. We can see a sketch of the curve below.

$y = \frac{ln~x}{x^2}$ 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 $y = 0$: $\frac{ln~x}{x^2} = 0$ $ln~x = 0$ $x = 1$ The x-intercept is $1$ C. The function is not an even function or an odd function. D. $\lim\limits_{x \to 0^+} \frac{ln~x}{x^2} = -\infty$ $x = 0$ is a vertical asymptote. $\lim\limits_{x \to \infty} \frac{ln~x}{x^2} =\frac{\infty}{\infty}$ $\lim\limits_{x \to \infty} \frac{\frac{1}{x}}{2x} = 0$ $y = 0$ is a horizontal asymptote. E. We can find the values of $x$ such that $y' = 0$: $y' = \frac{(\frac{1}{x})(x^2)~- (ln~x)(2x)}{x^4} = \frac{1-2~ln~x}{x^3} =0$ $1-2~ln~x = 0$ $ln~x = \frac{1}{2}$ $x = e^{1/2}$ $x = \sqrt{e}$ $x = 1.65$ When $0 \lt x \lt 1.65$, then $y' \gt 0$ The function is increasing on the interval $(0, 1.65)$ When $x \gt 1.65$, then $y' \lt 0$ The function is decreasing on the interval $(1.65, \infty)$ F. When $x = 1.65$, then $y = \frac{ln(1.65)}{(1.65)^2} = 0.184$ $(1.65,0.184)$ is a local maximum. G. We can find the values of $x$ such that $y'' = 0$: $y'' =\frac{(-\frac{2}{x})(x^3)-(1-2~ln~x)(3x^2)}{x^6}$ $y'' =\frac{(-2)-(1-2~ln~x)(3)}{x^4}$ $y'' =\frac{6~ln~x-5}{x^4} = 0$ $6~ln~x-5 = 0$ $ln~x = \frac{5}{6}$ $x = e^{5/6}$ $x = 2.3$ When $0 \lt x \lt 2.3$, then $y'' \lt 0$ The graph is concave down on the interval $(0, 2.3)$ When $x \gt 2.3$, then $y'' \gt 0$ The graph is concave up on the interval $(2.3, \infty)$ When $x = 2.3$, then $y = \frac{ln(2.3)}{(2.3)^2} = 0.157$ The point of inflection is $(2.3, 0.157)$ H. We can see a sketch of the curve below.