1. ## Extrema

Find All Extrema

y=(2x)/(x+4)^3

f '(x)= (x+4)^3(2)-(2x)(3)(x+4)(1)
-------------------------
((x+4)^3)^2)

Not sure what to do next. I know I can cancel out something. Can someone assist me?

2. Originally Posted by sderosa518
Find All Extrema

y=(2x)/(x+4)^3

f '(x)= (x+4)^3(2)-(2x)(3)(x+4)(1)
-------------------------
((x+4)^3)^2)

Not sure what to do next. I know I can cancel out something. Can someone assist me?
$f'(x)=\frac{(x+4)^3(2)-(2x)(3)(x+4)^2(1)}{((x+4)^3)^2}$

$=\frac{2(x+4)^3-6x(x+4)^2}{(x+4)^6}$

$=\frac{2(x+4)^2(x+4-3x)}{(x+4)^6}$

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

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

Set the numerator equal to 0 and solve for x to find the local extrema

3. So the extrema is 2??? Is there more extremas. Key Word ALL???

Correct

4(-x+2)
-------
(x+4)^4

4(-x+2)=0

4(-x+2)=0
------- --
4 4

(-x+2)=0

-x=-2

=2

4. Originally Posted by sderosa518
So the extrema is 2???
no. extrema are function values (y's).

x = 2 is a location for a possible function extrema.

conduct the 1st or 2nd derivative test for extrema to classify/confirm.

5. Originally Posted by sderosa518
So the extrema is 2??? Is there more extremas. Key Word ALL???

Correct

4(-x+2)
-------
(x+4)^4

4(-x+2)=0

4(-x+2)=0
------- --
4 4

(-x+2)=0

-x=-2

=2
For $x>2$, $f'(x)>0$

For $x<2$, $f'(x)<0$

$f'(x)$ changes from positive to negative at $x=2$

So $f(x)$ has a local max at $x=2$

To find the value of the local max, find $f(2)$

Yes, there is only 1 extremum

6. Originally Posted by sderosa518
Find All Extrema

y=(2x)/(x+4)^3

f '(x)= (x+4)^3(2)-(2x)(3)(x+4)(1)
-------------------------
((x+4)^3)^2)

Not sure what to do next. I know I can cancel out something. Can someone assist me?
Presumably you have used the quotient rule to find the derivative here. This just goes to show the futility of teaching the quotient rule (futile as such derivative can be found just as if not more easily using the product and chain rule).

Using the product rule:

$\frac{d}{dx}\left( \frac{2x}{(x+4)^3}\right)=\left(\frac{d}{dx}(2x)\r ight)\left( \frac{1}{(x+4)^3}\right)+(2x) \frac{d}{dx}\left( \frac{1}{(x+4)^3}\right)$ $=
\frac{2}{(x+4)^3}+\frac{2x(-3)}{(x+4)^4}=
\frac{4(2-x)}{(x+4)^4}$

CB