1. decreasing intervals

Find all open intervals on which the function $\displaystyle f(x) = \frac {x^2} {x^2 + 4}$ is decreasing

2. Originally Posted by Samantha
Find all open intervals on which the function $\displaystyle f(x) = \frac {x^2} {x^2 + 4}$ is decreasing
That is find the open intervals on which $\displaystyle \frac{d}{dx}\left[\frac{x^2}{x^2+4}\right] \le 0$

RonL

3. Originally Posted by CaptainBlack
That is find the open intervals on which $\displaystyle \frac{d}{dx}\left[\frac{x^2}{x^2+4}\right] \le 0$

RonL
Now:

$\displaystyle \frac{d}{dx}\left[\frac{x^2}{x^2+4}\right]=\frac{2x}{x^2+4}-\frac{2x^3}{(x^2+4)^2}$

which is less than or equal to zero when:

$\displaystyle 2x(x^2+4)-2x^3 \le 0$

RonL

4. Hello, Samantha!

. . You don't know the Quotient Rule?
. . You don't understand increasing/decreasing?
. . You can't handle inqualities?

Find all open intervals on which the function $\displaystyle f(x) \:=\: \frac {x^2} {x^2 + 4}$ is decreasing
Quotient Rule: .$\displaystyle f'(x) \;=\;\frac{(x^2+4)\cdot2x - x^2\cdot2x}{(x^2+4)^2}\;=\;\frac{8x}{(x^2+4)^2}$

The function is decreasing where $\displaystyle f'(x) \,<\,0$

. . So we have: .$\displaystyle \frac{8x}{(x^2+4)^2} \:<\:0$

Since the denominator is always positive, we have: .$\displaystyle 8x \:<\:0\quad\Rightarrow\quad x \:<\:0$

Therefore, $\displaystyle f(x)$ is decreasing on the interval: .$\displaystyle (-\infty,\:0)$

5. Originally Posted by Samantha
Find all open intervals on which the function $\displaystyle f(x) = \frac {x^2} {x^2 + 4}$ is decreasing
The curve also shows that f(x) is decreasing on (-infinity,0)