Derivative of graphs

Question

anonymous · Answer

attachment

tkhunny · Answer

Pick it apart and see who wins!

It needs asymptotes, but they all seem to have about the right ones.

It needs to be zero at x = 0.  You can discard A and C

It should head for zero toward both extremes.  They might all do that.  Not really clear.

Starting from the left, it should grow in the negative direction while approaching the first asymptote.  Discard D and E.

I think we ran out of choices.

anonymous · Answer

so it's B?

agent0smith · Answer

You can pretty quickly eliminate most... look at the left most portion of the orig. graph - the slope is almost flat, BUT it's sloping downwards... meaning a negative near zero slope. There is only two that match that - A and B

anonymous · Answer

1. trace the graph from left to right along x axis
2. if the curve increases, derivative should go positive, if decreases, it should be negative
3. at a minima or maxima, derivative must be "zero"
so, yes. B seems to be the one that makes sense.

agent0smith · Answer

B, C, D are all positive on the left most portion, meaning the graph would be sloping up, not down, so they're gone.

Then, of A and B, only one has a zero value (x-intercept) where it should. The other is always positive.

anonymous · Answer

okay got it thank you guys:)

tkhunny · Answer

I think what we might conclude is that these drawings aren't that good and there might be reasons to rule out all of them.  Tricky thing, drawings!

agent0smith · Answer

They aren't perfect, but B is pretty darn close. The x-intercept looks a bit off, though.

agent0smith · Answer

What i don't get is why is there all those random tickmarks above and to the side of the axes?? It looks like someone drew them here on openstudy |dw:1382742493550:dw|

tkhunny · Answer

I took a quick shot at it.

tkhunny · Answer

That's $f(x) = \dfrac{2}{(x+2)(1-x)}$

agent0smith · Answer

Looks about right.

tkhunny · Answer

...and my hash marks are FAR superior!  :-)

agent0smith · Answer

haha yeah. I think that may be the first set of graphs i've seen where the tick marks are not actually on the axes, instead they're just randomly placed near it. And not even consistently... some are higher than others.