f(x)=(x+3)/(x)^2

Question

f(x)=(x+3)/(x)^2

anonymous · Answer

$$f(x)=(x+3)/x^2$$ Find points of increase and decrease and all relative extrema.

anonymous · Answer

I found the derivative as $$(-x^2-6x)/(x^4)$$ I could take -x out of the top of the fraction... $$(-x(x+6))/(x^4)$$

anonymous · Answer

differentiate ie dy/dx=-(x+6)/x^3. for increasing let dy/dx >0 and decreasing dy/dx <0 and for extreme dy/dx=0

anonymous · Answer

I don'y understand how you got -(x+6)/x^3

anonymous · Answer

factor out an $x$ and cancel top and bottom

anonymous · Answer

you get
$$-\frac{x+6}{x^3}$$

anonymous · Answer

positive if $-6<x<0$ negative otherwise

anonymous · Answer

Okay, that's what I got too.

anonymous · Answer

I have areas of increase being (-6,0) and decrease being (-infinity, -6) (0, infinity)

anonymous · Answer

I'm not sure if that's correct. Also, my minimum and maximum do not match what the graph looks like.

anonymous · Answer

???

anonymous · Answer

ok let me try it
maybe the derivative is wrong

anonymous · Answer

no, it is rigth

anonymous · Answer

right

anonymous · Answer

okay, thank you

anonymous · Answer

decreasing on $(-\infty, -6)$ increasing on $(6,0)$ then decreasing on $(0,\infty)$ here is the picture http://www.wolframalpha.com/input/?i=y%3D%28x%2B3%29%2F%28x%29^2

anonymous · Answer

Then would there be a relative maximum at x=0?

anonymous · Answer

heck no!

anonymous · Answer

the function does not include 0 in its domain

anonymous · Answer

aka  you cannot divide by 0
that is why you see a vertical asymptote at $x=0$

anonymous · Answer

Okay, haha I understand.