Start by calculating the first derivative of #f(x)# - use the quotient rule
#d/dx(f(x)) = ( [d/dx(x^2)] * (x^2 + 3) - x^2 * d/dx(x^2 + 3))/(x^2 + 3)^2#
#f^' = (2x * (x^2 + 3) - x^2 * 2x)/(x^2 + 3)^2#
#f^' = (color(red)(cancel(color(black)(2x^3))) + 6x - color(red)(cancel(color(black)(2x^3))))/(x^2 + 3)^2 = (6x)/(x^2 + 3)^2#
Before calculating the second derivative, find the critical points of the function by having #f^' = 0#. These points will help you determine the local minimum and local maximum.
#(6x)/(x^2 + 3)^2 = 0 <=> 6x = 0 => x = color(green)(0) -># critical point
To determine where the function is increasing and where it's decreasing, examine the sign of the first derivative around the critical point #x=0#.
Since the numerator of #f^'# will always be positive, the sign of the first derivative will be determined by the numerator.
This means that you have #6x<0# for #x<0#, so the function is decreasing on #(-oo,0)#.
Likewise, because #6x>0# for #x>0#, the function is increasing on #(0, oo)#.
Now calculate the second derivative - use the quotient and the chain rules
#d/dx(f^') = ([d/dx(6x)] * (x^2 + 3)^2 - 6x * d/dx(x^2 + 3)^2)/[(x^2+3)^2]^2#
#f^('') = (6 * (x^2 + 3)^color(red)(cancel(color(black)(2))) - 6x * 2 * color(red)(cancel(color(black)((x^2+3)))) * 2x)/(x^2+3)^color(red)(cancel(color(black)(4)))#
#f^('') = (6 *(x^2 + 3) - 24x^2)/(x^2 + 3)^3#
#f^('') = (-18x^2 + 18)/(x^2 + 3)^3 = -(18 * (x^2-1))/(x^2 + 3)^3#
You can determine the local minimum and local maximum of the function by using the second derivative test, which tells you that if you have #c# as a critical point of #f(x)#, then the sign of #f^('')(c)# will tell you
So, your critical point is #c=0#, which means that you have
#f^('')(0) = -18 * (0^2 -1)/(0^2 + 3)^3#
#f^('')(0) = -18 * ((-1))/27 = 18/27 = 2/3#
The point #(0, f(0))# will be a local minimum. Moreover, the point #(0, f(0))# will be an absolute minimum as well, since
#f(x) = x^2/(x^2 + 3) > 0,(AA) x !=0# on #(-oo,oo)#
To determine where the function is concave up and where it's concave down, analyze the behavior of #f^('')# around the Inflection points, where #f^('')=0#.
#f^('') = -(18(x^2-1))/(x^2 + 3)^2=0#
This implies that
#-18(x^2-1) = 0 <=> x^2 = 1 => x = +-1#
Now you're going to see what the sign of the second derivative is for the intervals
It's clear that #f^('')# will be negative on this interval, so #f# will be concave down.
On this interval, #f^('')# will be positive, since #(x^2-1)# will be negative. As a result, #f# will be concave up.
Once again, #f^('')# will be negative for this interval, so #f# will be concave down.
