Question

For what intervals is

g(x) = 1
----------
x^2 + 1

concave down?

_____ < x < _________

please explain? thanks!!

Answer 1

so it would be

-∞ < x < ∞

? :/

Answer 2

@eliassaab your reply disappeared! :/

Answer 3

\[g(x)=\frac{1}{x^2+1}\\
g'(x)=-\frac{2x}{(x^2+1)^2}\\
\begin{align*}g''(x)&=\frac{-2(x^2+1)^2-(-2x)(2(x^2+1))(2x)}{(x^2+1)^4}\\&=\frac{-2(x^2+1)^2+8x^2(x^2+1)}{(x^2+1)^4}\\
&=\frac{(x^2+1)\left(-2(x^2+1)+8x^2\right)}{(x^2+1)^4}\\
&=\frac{6x^2-2}{(x^2+1)^3}\end{align*}\]
Find the critical points (when is \(g''(x)=0\) or undefined?).

Answer 4

Here is the answer it concave down for
\[
-\frac 1{\sqrt 3}\le x \le\frac 1{\sqrt 3}
\]

Answer 5

This where the second derivative is negative

Answer 6

so you simplified 6x^2-2/(x^2+1)^3 into + or - 1/√3

? :/ if so, could you please show me how to simplify that?

@SithsAndGiggles i'm following what you've written so far:)

Answer 7

The sigh depends on the numerator sign \[ 6x^2 -2\]

Answer 8

SIGN

Answer 9

Now
\[
6 x^2 -2 = 6( x - \frac1{\sqrt 3})( x + \frac1{\sqrt 3}) <0\\for \\
-\frac 1{\sqrt 3}\le x \le\frac 1{\sqrt 3}

\]

Answer 10

The denominator of the second derivative is always positive. So we need to study the sign of the numerator only.

Answer 11

Is it clear now?

Answer 12

yes :)

thanks so much!! :D

Answer 13

YW