# Chapter 4 - Section 4.3 - How Derivatives Affect the Shape of a Graph - 4.3 Exercises - Page 302: 52

(a) $x=0$ is a vertical asymptote. $y = 0$ is a horizontal asymptote. $y = -1$ is a horizontal asymptote. (b) $f$ is increasing on the intervals $(-\infty,0)\cup(0,\infty)$ (c) There is no local maximum or local minimum. (d) The graph is concave down on this interval: $(0, \infty)$ The graph is concave up on this interval: $(-\infty,0)$ There are no points of inflection. (e) We can see a sketch of the graph below. (a) $f(x) = \frac{e^x}{1-e^x}$ $f(x)$ is undefined when $1-e^x = 0$ $x=0$ is a vertical asymptote. $\lim\limits_{x \to -\infty}f(x) = 0$ $y = 0$ is a horizontal asymptote. $\lim\limits_{x \to \infty}f(x) = -1$ $y = -1$ is a horizontal asymptote. (b) We can find the points where $f'(x) = 0$: $f'(x) = \frac{(e^x)(1-e^x)-(e^x)(-e^x)}{(1-e^x)^2}$ $f'(x) = \frac{e^x-e^{2x}+e^{2x}}{(1-e^x)^2}$ $f'(x) = \frac{e^x}{(1-e^x)^2}$ There are no values of $x$ such that $f'(x) = 0$ Note that $f'(x)$ is undefined when $x=0$ $f$ is not decreasing on any interval. When $x \lt 0$ or $x \gt 0$ then $f'(x) \gt 0$ $f$ is increasing on the intervals $(-\infty,0)\cup(0,\infty)$ (c) There are no values of $x$ such that $f'(x) = 0$ There is no local maximum or local minimum. (d) We can find the points where $f''(x) = 0$: $f''(x) = \frac{(e^x)(1-e^x)^2-(e^x)(2)(1-e^x)(-e^x)}{(1-e^x)^4}$ $f''(x) = \frac{(e^x)(1-2e^x+e^{2x})+(2e^{2x})(1-e^x)}{(1-e^x)^4}$ $f''(x) = \frac{e^x-2e^{2x}+e^{3x}+2e^{2x}-2e^{3x}}{(1-e^x)^4}$ $f''(x) = \frac{e^x-e^{3x}}{(1-e^x)^4}$ $f''(x) = \frac{(1-e^x)(e^{2x}+e^x)}{(1-e^x)^4}$ $f''(x) = \frac{e^{2x}+e^x}{(1-e^x)^3}$ There are no values of $x$ such that $f''(x) = 0$ Note that $f''(x)$ is undefined when $x=0$ The graph is concave down when $f''(x) \lt 0$ The graph is concave down on this interval: $(0, \infty)$ The graph is concave up when $f''(x) \gt 0$ The graph is concave up on this interval: $(-\infty,0)$ There are no points of inflection. (e) We can see a sketch of the graph below. 