# graphing confusion of function and its derivative

• Apr 29th 2014, 12:03 AM
catenary
graphing confusion of function and its derivative
I was doing several function graphing calculus problems (already 31 problems) all with success and no confusion then suddenly I stumble upon this function:
--> f(x) = 2x+ (1/2x)
-->differentiation yielded f'(x) = 2 - (1/2x^2)
-->solving for the zeroes of the derivative yielded sqrt(1/4) which is plus and minus 0.5 which means that those places should be horizontal tangent points on the original function
-->But then i graph the original function on MathGV and get a straight line that look like this:
Attachment 30788
graph of 2x+(1/2x)
there are no horizontal tangent points!
-->So then I tried combining the original function into one fraction which looks like (4x^2+1)/2x and then graphed it and it looked like this:
Attachment 30789
graph of (4x^2+1)/2x
Now I am confused why it would look different as it is the same function. Also it seems there are still no horizontal tangent points.
-->So then I tried to graph the derivative the original function just to see if the zeroes that I found were correct and it looked like this:
Attachment 30790
graph of 2-(1/2x^2)
I am confused why the zeroes here are on plus and minus 2..
-->So as a final resort I decided to combine the derivative into one fraction which looks like (4x^2 - 1)/2x^2 and then graphed it:
Attachment 30791
graph of (4x^2+1)/2x
I am confused again why it would look different from the original derivative before it was combined into one fraction however this time the zeroes are correctly located on plus and minus 0.5

So how does the real graph of the original function 2x + (1/2x) look like?
Why would it look different when I combined it into one fraction?
Did I do something wrong or is there something I don't understand here?
Also how does the real graph of the derivative of the original function look like?
I expect the way you entered the function into your grapher has the computer reading the function as \displaystyle \begin{align*} 2x + \frac{1}{2}x \end{align*} instead of \displaystyle \begin{align*} 2x + \frac{1}{2x} \end{align*}. Try entering the function as y = 2*x + 1/(2*x).