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?
$\displaystyle f'(x)=\frac{(x+4)^3(2)-(2x)(3)(x+4)^2(1)}{((x+4)^3)^2}$
$\displaystyle =\frac{2(x+4)^3-6x(x+4)^2}{(x+4)^6}$
$\displaystyle =\frac{2(x+4)^2(x+4-3x)}{(x+4)^6}$
$\displaystyle =\frac{2(-2x+4)}{(x+4)^4}$
$\displaystyle =\frac{4(-x+2)}{(x+4)^4}$
Set the numerator equal to 0 and solve for x to find the local extrema
For $\displaystyle x>2$, $\displaystyle f'(x)>0$
For $\displaystyle x<2$, $\displaystyle f'(x)<0$
$\displaystyle f'(x)$ changes from positive to negative at $\displaystyle x=2$
So $\displaystyle f(x)$ has a local max at $\displaystyle x=2$
To find the value of the local max, find $\displaystyle f(2)$
Yes, there is only 1 extremum
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:
$\displaystyle \frac{d}{dx}\left( \frac{2x}{(x+4)^3}\right)=\left(\frac{d}{dx}(2x)\r ight)\left( \frac{1}{(x+4)^3}\right)+(2x) \frac{d}{dx}\left( \frac{1}{(x+4)^3}\right)$ $\displaystyle =
\frac{2}{(x+4)^3}+\frac{2x(-3)}{(x+4)^4}=
\frac{4(2-x)}{(x+4)^4}$
CB