$$f(x)=\ln \left|\frac{x}{1+x^2}\right|$$
I am told the range of f(x) is: $f(x)\le \ln\frac{1}{2}=-\ln2$ but don't know how to get there.
I do know $f(x)$ becomes a piecewise function b/c of the absolute value based on $x>0$ and $x<0$ (Undefined at $x=0$) I also know that the domain of any $\ln(f(x))$ is that $f(x)\gt0$. I'm just not sure how to get the range.
$\endgroup$ 63 Answers
$\begingroup$As Antonio Vargas said, it will help to first find the range of $M = |\frac x {1 + x^2}|$.
Since an absolute value is never negative, if $M$ can be zero then zero is the minimum. I believe you can find the value of $x$ for which $M$ is zero.
Now we need to find out the maximum. How large can $M$ get? Notice that the denominator is "bigger" than the numerator, so for large values of $x$, $M$ approaches zero. That is: $$\lim_{x \rightarrow \infty} M = 0$$
So somewhere before $x = \infty$, $M$ has a maximum value. Since you tagged this as calculus you probably know to take a derivative to find a max:
$$\frac{d}{dx}\frac{x}{1 + x^2} = \frac{(1 - x)(1 + x)}{\left(x^2+1\right)^2} = 0$$
So your max will occur at $x=1$ or $x=-1$. Now can you find the range of $M$, and thus the range of $\ln M$?
$\endgroup$ 8 $\begingroup$Here is how I did it. Here is how I did it:
Hint By AM-GM, $\dfrac{1+x^2}2 \ge \lvert x \rvert$.
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