1. ## Max/Min

Let f be the function defined by f(x) = ln absolute value of $\frac{x}{1+ x ^2}$

b) Determine whether f is an even function, odd function or neither. Justify.

c) At what values does f have relative maximum or minimum? Use first derivative test to determine whether f(x) is a relative max/ min.

2. Originally Posted by ny_chow
Let f be the function defined by $f(x)=\left|\frac{x}{1+ x ^2}\right|$

b) Determine whether f is an even function, odd function or neither. Justify.

c) At what values does f have relative maximum or minimum? Use first derivative test to determine whether f(x) is a relative max/ min.
b) Well, $1+x^2>0$ for all x, so $f(x) = \frac{|x|}{1+x^2}$

To be even, $f(x)=f(-x)$, which is true here, because $|x|$ is even, $1+x^2$ is even, and the composition of even functions is even.

c) $f'(x) = \frac{(x^2+1)\cdot sign(x)-|x|(2x)}{(1+x^2)^2} = \frac{(x^2+1)\cdot sign(x)-(2x^2)\cdot sign(x)}{(1+x^2)^2}$ $= sign(x)\cdot\frac{x^2+1-2x^2}{(1+x^2)^2} = sign(x)\cdot\frac{1-x^2}{(1+x^2)^2}$

Solving $f'(x)=0$ gives $x=\pm 1$.

Taking the second derivative (just use the quotient rule and treat $sign(x)$ as a constant) gives you $sign(x)\cdot\frac{2x(x^2-3)}{(1+x^2)^3}$. Plugging in $x=1$ and $x=-1$ both give you negative values so they are both local maximums.

It is also worth noting that the $sign(0)=0$ so $f'(0)=0$. This is a local minimum as zero is the unique point for which $f(x)=0$; at all other values $f(x)>0$.

3. what does
$sign (x)$ mean?

4. Originally Posted by ny_chow
what does
$sign (x)$ mean?
It just tells you the sign of the number. It's explicit formula is:
$
sign(x)=\left\{\begin{array}{lr}1&:x>0\\0&:x=0\\-1&:x<0\end{array}\right\}$