Expert: Steve Holleran - 9/16/2007

Question
Hi,

How would you find the stationary points of the function   f(x)= 3+4x/1+x^2 which when using the Quotient rule has the deriative
f(x)=2(2-3x-2x^2)/(1+x2)^2

I want to use the First Deriative test to find each
stationary point as a local maximum or a local minimum of f(x)?

Answer
Hello Alexcia,

Okay, all you want to do is set your first derivative = 0

The part of the numerator in the parentheses can be factored:

  -2x^2 -3x + 2 = -1(2x^2 + 3x - 2)

                = -1(2x - 1)(x + 2)

So then the derivative will = 0 when each of these factors are =0, so that occurs at x = 1/2 and x = -2.

I tell students the easiest way to look at the type of extrema you have is with a sign chart of the first derivative:  

 f'(x) = [-2 * (2x-1) * (x+2)] / [(1+x^2)^2]

The split points on the sign chart are where the top or bottom are = 0. But here, the bottom is never = 0 , so just use the stationary points we found.  The chart looks like:

f'  -----------|-----------------------|-------------
         -               +                     -
             -2                       1/2

So, since f' changes from - to + at x = -2, the function goes from decreasing to increasing, and you have a local min; and since it goes from + to - as x = 1/2, the function goes from increasing to decreasing, and you have a local max.

(in case you are unfamiliar with getting the sign chart, all you do is pick "test values" of x in each of the intervals created by the sign chart, and see what the sign of f' would be using those test values.  I used x = -3 for the interval to the left of -2 ; 0 for the interval between -2 and 1/2 ; and 1 for the interval to the right of 1/2.

Hope this helps

Steve Holleran