Answer to Question #332078 in Calculus for LM.

Question

1. Let f be the function defined by

f (x) = x2

-------------

(2x + 1)2

:

(a) Determine the vertical and horizontal asymptotes (show all limits). (4)

(b) Use the sign pattern for f ' (x) to determine

(i) the interval(s) over which f rises and where it falls; (4)

(ii) the local extrema. (2)

(c) Use the sign pattern for f " (x) to determine

(i) where the graph of f is concave up and where it is concave down. (4)

(ii) the inflection points (if any) (2)

Answer 1

$f(x)=\frac{x^2}{(2x+1)^2}$ , $x\neq -\frac{1}2$

(a) Vertical asymptotes: $(2x+1)^2=0,\quad 2x+1=0,\quad x=-\frac{1}{2}$ .

$\lim\limits_{x\rightarrow -1/2}\frac{x^2}{(2x+1)^2}=\infty ,\quad$ $x=-\frac{1}{2}$ is a vertical asymptote.

Horizontal asymptotes: $\lim\limits_{x\rightarrow\infty}\frac{x^2}{(2x+1)^2}= \lim\limits_{x\rightarrow\infty}\frac{x^2}{4x^2+4x+1}= \lim\limits_{x\rightarrow\infty}\frac{1}{4+4/x+1/x^2}=\frac{1}{4}$

$y=\frac{1}{4}$ is a horizontal asymptote.

(b) $f^\prime (x)=\frac{2x\cdot (2x+1)^2-4(2x+1)\cdot x^2}{(2x+1)^2}=\frac{2x}{(2x+1)^3}$

(i) If $-\tfrac{1}{2}<x<0$ , then $f^\prime (x)<0$ and $f(x)$ falls.

If $x<-\frac{1}{2}$ or $x\geq 0$ , then $f^\prime (x)>0$ and $f(x)$ rises.

(ii) $f(x)=0$ , $\frac{2x}{(2x+1)^3}=0$ , $x=0$ .

The point $x=0$ is a local minimum (if $x<0$ , then $f$ falls; if $x>0$ , then $f$ rises).

(c) $f^{\prime \prime}(x)=\frac{2\cdot (2x+1)^3-6(2x+1)^2\cdot 2x}{(2x+1)^6}=\frac{2-8x}{(2x+1)^4}$

(i) If $x<\frac{1}{4}$ , then $f^{\prime\prime}(x)>0$ and $f(x)$ is concave up.

If $x>\frac{1}{4}$ , then $f^{\prime\prime}(x)<0$ and $f(x)$ is concave down.

(ii) $f^{\prime\prime}(x)=0,\ \ \frac{2-8x}{(2x+1)^4}=0,\ \ x=\frac{1}{4}$ .

The point $x=\frac{1}{4}$ is an inflection point (if $x<\frac{1}{4}$ , then $f$ is concave up; if $x>\frac{1}{4}$ , then $f$ is concave down).