Need help on graphing this function.

[jzq]

Sketch the graph of the function, using the curve-sketching guide.

Function: \frac {x}{x^2-4}

So far I have derived this information from the function: (Please check!)

Domain: (-\infty,-2)\cup(-2,2)\cup(2,\infty)
y-int: (0,0)
x-int: (0,0)
Asymptote: x=-2 , x=2
First Derivative: f'(x)=\frac {-x^2-4}{(x^2-4)^2}
Second Derivative: f''(x)=\frac {2x(x^2+12)}{(x^2-4)^3}

The information that I need now is where it is increasing and decreasing; the relative minimum; where it concaves up and down; and the points of inflection. My problem is, when I plug in zero for y in the derivatives, it gets complicated. For example, the first derivative: (Please check!)

\frac {-x^2-4}{(x^2-4)^2}=0

-x^2-4=0

-x^2=4

x^2=-4

x=\sqrt{-4}

Correct me if I'm wrong, if you square root a negative number, wouldn't you have to use imaginary numbers (\imath) ? Any help will be greatly appreciated!

[Davorak]

The derivative method will not necessarily work when the function has asymptotes like this one. Find the asymptotes and then find the concavity for each area. Plot it out by find if it does not make sense.
Here is online graphing java applet to help out.
http://www.math.unl.edu/~jorr/java/html/Grapher.html

[ehild]

Sketch the graph of the function, using the curve-sketching guide.

Function: \frac {x}{x^2-4}

So far I have derived this information from the function: (Please check!)

First Derivative: f'(x)=\frac {-x^2-4}{(x^2-4)^2}
Second Derivative: f''(x)=\frac {2x(x^2+12)}{(x^2-4)^3}

The information that I need now is where it is increasing and decreasing; the relative minimum; where it concaves up and down; and the points of inflection. My problem is, when I plug in zero for y in the derivatives, it gets complicated.



You should stay in the real world, no complex numbers are needed. See f ' (x). It is always negative, it cannot be zero or positive. So your function is decreasing everywhere. It has neither local maxima nor minima.
f '' (x) is zero at x=0, and there are domains where it is either positive or negative, so the function has got an inflexion point and domains where it is concave up or down.


ehild

[dextercioby]

Here is (or at least should be) your graph...


Daniel.

[jzq]

The derivative method will not necessarily work when the function has asymptotes like this one. Find the asymptotes and then find the concavity for each area. Plot it out by find if it does not make sense.
Here is online graphing java applet to help out.
http://www.math.unl.edu/~jorr/java/html/Grapher.html
The way we're learning it, we're suppose to draw the graph w/o copying from a graphing calculator using the information derived from the function. After we found the derivatives, we have to set it = to 0, then find the critical numbers. With the critical numbers, we're then suppose to test it on a number line to find where it is positive or negative. From the first derivative number line, we find the decreasing and increasing intervals and the relative extrema. From the second derivative number line, we find the intervals where it concaves up and down and the points of inflection. I'm sure you all know this already. So basically, on my homework, I have to show how I got all my information. Thanks for your help!