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$$f(x)=\ln \left|\frac{x}{1+x^2}\right|$$

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.

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First find the range of $\left|x/(1+x^2)\right|$, then use this to find the range of $f(x)$. –  Antonio Vargas Mar 10 at 13:33
    
OK, but I haven't found the range of a function in some time. Can you point me in the right direction? Domain is all Reals. So, is range all reals also? With the || added, range is y>=0. So range of ln|| would be ??? –  JackOfAll Mar 10 at 13:37
    
To simplify your life, you can remark that $|\frac{x}{1+x^2}|$ is an even function, so you can study it for $x>0$. On $(0,\infty)$, we have $|\frac{x}{1+x^2}|=\frac{x}{1+x^2}$. –  Taladris Mar 10 at 13:42
    
No, Jack, the range is not all reals also. Remember, the range of a function is the set of all of its possible outputs. Take a look at this plot, where the function $|x/(1+x^2)|$ is drawn in blue. The red lines are the boundary of its range. No matter what the $x$ value is, $|x/(1+x^2)|$ will always be between the top red line and the bottom red line. i.stack.imgur.com/AUKtH.png Now, where are the red lines? From the graph it sure looks like they're at $0$ and $1/2$, but how do you know for certain? –  Antonio Vargas Mar 10 at 13:44
    
Antonio, I need to do this without a graphing utility. –  JackOfAll Mar 10 at 15:20
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3 Answers 3

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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 $\ln M$?

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This was a big help. Thanks! –  JackOfAll Mar 10 at 15:36
    
What about the minumum value of the range? Are we assuming the lower end of the range is simply 0 b/c of the ln()? –  JackOfAll Mar 10 at 16:15
    
Can you figure out what the smallest value M can be is? –  DanielV Mar 10 at 16:20
    
No, that's not clear. I know if you have log(x) then x>0 since x is the results of an exponent operation. Otherwise, there are no critical points that are mins. And it's even. And the lim x-> infinity is 0. Doesn't this all point to the lowest value being y=0? Clearly, that's not right. –  JackOfAll Mar 12 at 1:35
    
You need to first figure out what the smallest value of $M$ can be. $M$ does not have a log function. –  DanielV Mar 12 at 6:28
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Here is how I did it. Here is how I did it:

enter image description here

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Well done, I'll point out that the statements $y \le -\ln 2$ and "the maximum value of $y$ is $-\ln 2$" are not exactly the same. Now you just need to find a lower bound. –  DanielV Mar 10 at 16:24
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Hint By AM-GM, $\dfrac{1+x^2}2 \ge \lvert x \rvert$.

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What is AM-GM ? –  JackOfAll Mar 10 at 13:40
    
Arithmetic mean - Geometric mean –  Max Mar 10 at 13:41
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