Hello everyone, I'm back again....haha. I was just checking over my homework assignment again and realized that I'm pretty unsure of one of the assigned problems. Here it is: Find the Intervals of Increase and Decrease, local max and min values and the concavity of the function f(x)= (x^2)/(x^2+3) First to find the intervals of increase and decrease as well as the local max and min values I found the first derivative of the function: f'(x)= [(x^2+3)(x)-(x^2)(2x)]/_x^2+3)^2 f ' (x) = (6x)/(x^2+3)^2 If you find where x equals zero you get: 6x=0 therefore x=0 x^2+3=0 therefore x=squareroot (-3) Now what I did was I went on to make a chart showing where the function was increasing and decreasing which I am unable to show on the computer and from that I got that it was increasing on the interval (0,infinity) and decreasing on the interval (-infinity,0). MY problem with this is that I just realized that to do this I took the square root of a negative number, which of course you cannot do. So do I just say that this is an unreal answer? Then using the chart I made I said that a local minimum occurs at (0,0) and that a local maximum does not occur for this graph. Then to find the concavity I found the second derivative of the function: f''(x)= [(x^2+3)^2(6)-24x^2(x^2+3)]/(x^2+3)^4 f ''(x) = [-18x(x-1)]/(x^2+3)^3 Then once again setting x=0 you get x=0, x=1, and then you run into the taking the square root of a negative number again. Which I never noticed at the time and I showed it as +square root 3 and -square root 3. Then I made another chart and showed where the graph was concave up or down and where the inflection points were. When I did this I got it was: Concave down (-infinity,0) U (1,infinity) Concave Up (0,1) Inflection points at x=0 and x=1. I thought I did it right until I was looking through my answers just now and noticed I was taking the root of a negative number, and now I'm not really sure what to do. Thanks in advance for any advice.