# min/max and concavity

• Jun 4th 2006, 05:38 PM
pakman
min/max and concavity
The question asks to find increasing/decreasing/concave up/concave down for:
(5x)/(x^2+1)

So I found the first derivative using the quotient rule:
[5(x^2+1)-5x(2x)]/[(x^2+1)^2]

Simplified, I got it down to
[-5x^2+5]\[(x^2+1)^2]

Got a critical point at x=1, and did a first derivate sign test to find increasing/decreasing. Came out to increasing: (-i, 1] and decreasing: [1, i)

On to concavity, which is found through the second derivative. This is where I am stumped. The algebra is killing me, or I could've messed up on the first derivate. Anyway, without simplification I got:
[-10x(x^2+1)^2]-[(-5x^2+5)(2(x^2+1)(2x))]

Could someone help me simplify that, it's killing me :mad:

• Jun 4th 2006, 06:23 PM
ThePerfectHacker
Quote:

Originally Posted by pakman
The question asks to find increasing/decreasing/concave up/concave down for:
(5x)/(x^2+1)

You have,
$y=\frac{5x}{x^2+1}$
Then,
$y'=\frac{(5x)'(x^2+1)-(5x)(x^2+1)'}{(x^2+1)}$
Thus,
$y'=\frac{5(x^2+1)-5x(2x)}{(x^2+1)^2}$
Thus,
$y'=\frac{5-5x^2}{(x^2+1)^2}$
---
Find the critical points, since the derivative exists everywhere you need to find,
$\frac{5-5x^2}{(x^2+1)^2}=0$
That happens when,
$5-5x^2=0$
Thus, $x=-1,1$.

Now, break your problem into intervals (you are going to use the first derivative test).

Considers the intervals, by choosing and point and calculating is derivative and seeing whether positive or negative.
$\left\{ \begin{array}{cc}x<-1&\mbox{ negative}\\-11& \mbox{ negative}$
Therefore the function is
Decreasing on $x<-1 \mbox{ and }x>1$
Inscreasing on $-1.
Also, from this test we see that a relative maximum occurs at $x=1$ and relative minimum at $x=-1$.
---
To work with concavity you need to find $y''$ which is by the quotient rule,
$y''=\frac{(5-5x^2)'[(x^2+1)^2]-(5-5x^2)[(x^2+1)^2]'}{(x^2+1)^4}$
Thus,
$y''=\frac{(-10x)[(x^2+1)^2]-2x(5-5x^2)(x^2+1)}{(x^2+1)^4}$
Thus,
$y''=\frac{(x^2+1)[(-10x)(x^2+1)-2x(5-5x^2)]}{(x^2+1)^4}$
Thus,
$y''=\frac{(-10x)(x^2+1)-2x(5-5x^2)}{(x^2+1)^3}$
Thus,
$y''=\frac{-10x^3-10x-10x+10x^3}{(x^2+1)^3}$
Thus,
$y''=\frac{-20x}{(x^2+1)^3}$

Proceed as before by finding the critical points,
$\frac{-20x}{(x^2+1)^3}=0$
Thus,
$-20x=0$ thus, $x=0$.

Now divide the intervals into two parts,
$x<0$ and $x>0$ select any points and these intervals and find thier signs.
$\left\{ \begin{array}{cc}x<0 \mbox{ positive}\\ x>0 \mbox{ negative}$
Therefore,
$x<0$ concave down,
$x>0$ concave up.
• Jun 5th 2006, 11:34 PM
earboth
Quote:

Originally Posted by ThePerfectHacker
...
To work with concavity you need to find $y''$ which is by the quotient rule,
$y''=\frac{(5-5x^2)'[(x^2+1)^2]-(5-5x^2)[(x^2+1)^2]'}{(x^2+1)^4}$
Thus,
$y''=\frac{(-10x)[(x^2+1)^2]-2x(5-5x^2)(x^2+1)}{(x^2+1)^4}$
...

Hello,

If you look at the graph of this function, you may have noticed that your result can't be correct:

$y''=\frac{(-10x)[(x^2+1)^2]-2 \cdot 2x(5-5x^2)(x^2+1)}{(x^2+1)^4}$

If you derivate $[(x^2+1)^2]'=2\cdot 2x\cdot (x^2+1)$

$y''=\frac{10x(x^2-3)}{(x^2+1)^3}$