Guided Example #3 - Curve Sketching II

Sketch the graph of ${\displaystyle y=\frac{x^2}{x^2-1}}$

Step 1

Let's use the guidelines:
1) The domain is ${\displaystyle \{x | x\ne \pm 1\} \cup (-\infty ,-1)(-1,1) \cup (1,\infty )}$.

Step 2

2) The x- and y-intercepts are both 0.

Step 3

3) $\lim_{x\to \pm\infty}\frac{x^2}{x^2-1}=\lim_{x\to\pm\infty}\frac{1}{1-\frac{1}{x^2}}=1$
Therefore, the line ${\displaystyle y=1}$ is a horizontal asymptote.
Since the denominator is 0 when x= ±1, we compute the following limits:
$\lim_{x\to 1^+}\frac{x^2}{x^2-1}=\infty$       $\lim_{x\to 1^-}\frac{x^2}{x^2-1}=-\infty$
$\lim_{x\to -1^+}\frac{x^2}{x^2-1}=-\infty$     $\lim_{x\to -1^-}\frac{x^2}{x^2-1}=-\infty$
Therefore, the lines x = 1 and x = -1 are vertical asymptotes.

Step 4

4) $f\prime(x)=\frac{2x(x^2-1)-x^2(2x)}{(x^2-1)^2}=\frac{-2x}{(x^2-1)^2}$
We can see that ${\displaystyle f^{\text{'}}\left(x\right)>0}$ when and when ${\displaystyle x>0 (x\ne 1)}$.
Therefore, f is increasing on ${\displaystyle (-\infty ,-1)}$and ${\displaystyle (-1,0)}$ and decreasing on ${\displaystyle (0,1)}$ and ${\displaystyle (1,\infty )}$.

Step 5

5) Since ${\displaystyle f^{\text{'}}}$ changes from positive to negative at 0, there is a local maximum at ${\displaystyle (0,0)}$.

Step 6

6) $f\prime\prime(x)=\frac{-2(x^2-1)^2+(2x)(2)(x^2-1)(2x)}{(x^2-1)^4}=\frac{2(3x^2+1)}{(x^2-1)^3}$
Since ${\displaystyle 3x^2+1>0}$ for all x, we have
${\displaystyle f}$" ${\displaystyle \left(x\right)>0\leftrightarrow x^2-1>0\leftrightarrow \left|x\right|>1}$ and similarly ${\displaystyle f}$'' .
Hence, the graph is concave upward on the intervals ${\displaystyle (-\infty ,-1)}$and ${\displaystyle (1,\infty )}$ and concave downward on ${\displaystyle (-1,1)}$.
Note that there are no points of inflection since 1 and -1 are not in the domain.

Step 7

7) Using the information above we sketch the graph of f.

All Steps

Let's use the guidelines:
1) The domain is ${\displaystyle \{x | x\ne \pm 1\} \cup (-\infty ,-1)(-1,1) \cup (1,\infty )}$.

2) The x- and y-intercepts are both 0.

3) $\lim_{x\to \pm\infty}\frac{x^2}{x^2-1}=\lim_{x\to\pm\infty}\frac{1}{1-\frac{1}{x^2}}=1$
Therefore, the line ${\displaystyle y=1}$ is a horizontal asymptote.
Since the denominator is 0 when x= ±1, we compute the following limits:
$\lim_{x\to 1^+}\frac{x^2}{x^2-1}=\infty$       $\lim_{x\to 1^-}\frac{x^2}{x^2-1}=-\infty$
$\lim_{x\to -1^+}\frac{x^2}{x^2-1}=-\infty$     $\lim_{x\to -1^-}\frac{x^2}{x^2-1}=-\infty$
Therefore, the lines x = 1 and x = -1 are vertical asymptotes.

4) $f\prime(x)=\frac{2x(x^2-1)-x^2(2x)}{(x^2-1)^2}=\frac{-2x}{(x^2-1)^2}$
We can see that ${\displaystyle f^{\text{'}}\left(x\right)>0}$ when and when ${\displaystyle x>0 (x\ne 1)}$.
Therefore, f is increasing on ${\displaystyle (-\infty ,-1)}$and ${\displaystyle (-1,0)}$ and decreasing on ${\displaystyle (0,1)}$ and ${\displaystyle (1,\infty )}$.

5) Since ${\displaystyle f^{\text{'}}}$ changes from positive to negative at 0, there is a local maximum at ${\displaystyle (0,0)}$.

6) $f\prime\prime(x)=\frac{-2(x^2-1)^2+(2x)(2)(x^2-1)(2x)}{(x^2-1)^4}=\frac{2(3x^2+1)}{(x^2-1)^3}$
Since ${\displaystyle 3x^2+1>0}$ for all x, we have
${\displaystyle f}$" ${\displaystyle \left(x\right)>0\leftrightarrow x^2-1>0\leftrightarrow \left|x\right|>1}$ and similarly ${\displaystyle f}$'' .
Hence, the graph is concave upward on the intervals ${\displaystyle (-\infty ,-1)}$and ${\displaystyle (1,\infty )}$ and concave downward on ${\displaystyle (-1,1)}$.
Note that there are no points of inflection since 1 and -1 are not in the domain.

7) Using the information above we sketch the graph of f.