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$f(x) ={\frac{2}{1+x^2}}$

I need to find where this function is increasing, decreasing, concave and convex. I've found it's derivative:

$f'(x)=\frac{-4x}{(1+x^2)^2}$

Now you're supposed to make either $f'(x)>0$ when it's increasing and $f'(x)<0$ when it's decreasing, but that gives:

Increasing: $x<0$ Decreasing: $x>0$

But what does that actually mean? It's just confusing, usually when I solve these you get 2 solutions, so it's for example increasing on the interval of $(-2,2)$. What does this one tell me? What's the easiest way to find where this function is increasing and decreasing?

Then I also did the second derivative, which is:

$f''(x)= \frac{4(3x^2-1)}{(1+x^2)^3}$

How does this all help me find my solution?

peroxy
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  • The derivative is the slope of the graph. It tells how much the value changes in the y direction as a result of change in the x axis. When you have a negative derivative, it means that for a larger value of x you will have a smaller value for y. The opposite is true for a positive derivative. – PhzksStdnt Dec 03 '14 at 21:33
  • You've actually done it correctly, and the function is increasing for $x<0$ and decreasing for $x>0$ - try graphing it if you want to be sure. Do you know how to find where it's concave and convex though? – mardat Dec 03 '14 at 21:33

1 Answers1

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  1. The endpoints $\pm\infty$ are always present. Your results say that $f$ is increasing on $(-\infty, 0)$ and decreasing on $(0, \infty)$.
  2. For convexity / concavity you need to do the same thing to $f''$, $f'' > 0$ implies convexity and $f'' < 0$ implies concavity. Note that the numerator is the only interesing component of $f''$ because the denominator is always $\ge 1$, so it doesn't change the sign. Now to find the critical points, solve for $$4(3x^2-1) = 0 \Leftrightarrow x^2-\frac13=0 \Leftrightarrow x = \pm \frac1{\sqrt3}$$
AlexR
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  • Wait, so does that mean it's convex or concave? Or does that mean it's convex in $x=\frac{1}{\sqrt3}$ and concave in $x=-\frac{1}{\sqrt3}$? – peroxy Dec 03 '14 at 21:45
  • @NejcPisk You will have to look at three intervals: $(-\infty, -\frac1{\sqrt3})$, $(-\frac1{\sqrt3}, \frac1{\sqrt3})$ and $(\frac1{\sqrt3}, \infty)$. Intuition tells me that $f$ will be concave in the middle interval and convex on the outers. – AlexR Dec 03 '14 at 21:46
  • So I basically have to insert 3 values from those intervals into $f''(x)$? For example, $f''(10)>0$, does that mean it's convex? Or am I doing this the wrong way? – peroxy Dec 03 '14 at 21:57
  • @NejcPisk That's okay and shows that $f$ is convex on $(\frac1{\sqrt3}, \infty)$. Proceed the same way with the other intervals. – AlexR Dec 03 '14 at 21:58
  • Great! Thanks for explanation, this will be helpful during finals tomorrow! :) – peroxy Dec 03 '14 at 21:58