How do you find the points where the graph of the function # F(x)=x/(x^2+4)# has horizontal tangents and what is the equation?

1 Answer
Mar 28, 2016

The tangent is horizontal at #(2, 1/4)# and #(-2, -1/4)#.
The equations of these tangents parallel to x-axis are #y=+-1/4#.

Explanation:

y = #x/(x^2+4)#
#y' = (4-x^2)/(x^2+4)^2# = 0, when x = #+-2#.
The points at which the tangents are horizontal are #(2, 1/4)# and #(-2,-1/4).#
The equations of 0-slope horizontal tangents are y=#+-1/4.#

The graph passes through the origin, with y increasing between the turning points #(-2, -1/4)# and #(2, 1/4)# and decreasing to the limit 0, outside #-2<=x<2#, as #xto+-oo#..