Question

1. the function ( f ) and its first and second derivatives are given:

( f(x)=\frac{x}{x^{2}-1}, quad f^{prime}(x)=\frac{-left(x^{2}+1
ight)}{left(x^{2}-1
ight)^{2}}, quad f^{prime prime}(x)=\frac{2 xleft(x^{2}+3
ight)}{left(x^{2}-1
ight)^{3}} )
obtain the following information. write none when appropriate.
(a) ( x )-intercepts: , ( y )-intercepts:
(b) vertical asymptotes: horizontal asymptotes:
(c) increasing on: , decreasing on:

Explanation:

Step1: Find \(x\) - intercepts

Set \(y = f(x)=0\), i.e., \(\frac{x}{x^{2}-1}=0\). Since the numerator \(x = 0\) (and \(x^{2}-1
eq0\) when \(x = 0\)), the \(x\) - intercept is \(x = 0\).

Step2: Find \(y\) - intercepts

Set \(x = 0\) in \(y=f(x)\). Then \(y=\frac{0}{0^{2}-1}=0\), so the \(y\) - intercept is \(y = 0\).

Step3: Find vertical asymptotes

Set the denominator of \(f(x)\) equal to zero: \(x^{2}-1=(x + 1)(x - 1)=0\). So \(x=-1\) and \(x = 1\) are vertical asymptotes.

Step4: Find horizontal asymptotes

Use the limit \(\lim_{x
ightarrow\pm\infty}f(x)=\lim_{x
ightarrow\pm\infty}\frac{x}{x^{2}-1}=\lim_{x
ightarrow\pm\infty}\frac{\frac{x}{x^{2}}}{\frac{x^{2}}{x^{2}}-\frac{1}{x^{2}}}=\lim_{x
ightarrow\pm\infty}\frac{\frac{1}{x}}{1-\frac{1}{x^{2}}}=0\). So \(y = 0\) is a horizontal asymptote.

Step5: Determine intervals of increase and decrease

The first derivative \(f^{\prime}(x)=\frac{-(x^{2}+1)}{(x^{2}-1)^{2}}\). Since \(x^{2}+1>0\) and \((x^{2}-1)^{2}>0\) for \(x
eq\pm1\), \(f^{\prime}(x)<0\) for all \(x
eq\pm1\). The function is decreasing on \((-\infty,-1)\), \((-1,1)\) and \((1,\infty)\) and there is no interval where the function is increasing.

Answer:

(a) \(x\) - intercepts: \(0\), \(y\) - intercepts: \(0\)
(b) Vertical asymptotes: \(x=-1,x = 1\), Horizontal asymptotes: \(y = 0\)
(c) Increasing on: NONE, decreasing on: \((-\infty,-1)\cup(-1,1)\cup(1,\infty)\)