# Intervals of Increase/Decrease, and concavity

1. Nov 13, 2005

### scorpa

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.

2. Nov 13, 2005

### scorpa

Hmmm even if you use the values of -root3 and +root3 I think the answer still stays the same.

3. Nov 14, 2005

### hypermorphism

The places where this function blows up do not occur for any real value (if this were a function of complex variables, then you could consider complex roots). Of course, you should be able to see this without having to look at the derivative.
Now, about the first derivative, note that the denominator is irrelevant with respect to the sign of the derivative, and it is never 0 so you don't have to worry about invalid inputs.
Thus, you can just look at how 6x behaves in order to see where the graph is increasing and decreasing, which immediately gives you the intervals you describe.
Inflection points are not simply where the second derivative is 0. They are points where the graph changes concavity, which implies that if a point is an inflection point, then the second derivative is 0 there. The reverse is not true. For example, take f(x) = x10. The second derivative (as well as the 3rd, 4th, and many others) is zero at x=0, but there is no inflection point there. The graph is just "very flat" there.
In your problem, you already know that (0, f(0)) is a minimum; it cannot also be a point of inflection.

Last edited: Nov 14, 2005
4. Nov 14, 2005

### HallsofIvy

The point is that in $f'(x)= \frac{6x}{(x^2+3)^2}$
the denominator is always positive. The derivative is positive if and only if the numerator is positive: x> 0, and negative if and only if the numerator is negative: x< 0.

5. Nov 14, 2005

### scorpa

Ok so ignore the denominator, and just look at the 6x I will give that a try thank you!!!

6. Nov 14, 2005

### scorpa

OK so I redid the question and caught the error in my second derivative. I have now found the entire graph to be concave downwards and have no inflection points. My new second derivative is f''(x)= [18(1-x^2)]/(x^2+3)^3
Have I finally managed to fix the question?

7. Nov 14, 2005

### hypermorphism

How did you manage to show that ? 1-x2 is an inverted parabola raised by 1 unit in the y-axis and thus crosses the x-axis twice. Also note that the second derivative is positive at x=0, which implies the graph is concave upwards there (the first derivative is increasing locally; look at tangent lines as their slope increases, or just think of the second derivative of a standard concave up parabola).

8. Nov 14, 2005

### scorpa

Lol yeah my mind doesn't function well at 6 in the morning I guess. I had a few minutes before my 8 am class and i caught my mistake. Thanks a lot for the help, I appreciate it.