How do you find the points where the graph of the function #y = (x^3) + x# has horizontal tangents and what is the equation?

Question

1427 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-22T02:22:12

#y# has no real points through which there are horizontal tangents

#y=x^3+x#

#y# would have horizontal tangents at points where #y'(x) =0# - if such points exist for #x in RR#

#y'(x) = 3x^2+1#

#y'(x) = 0 -> 3x^2+1=0#

#x=sqrt(-1/3) = +-1/sqrt3i#

Hence: #y'(x)=0# has no #x in RR#

This can be seen by looking at the graph of #y# below:

graph{x^3+x [-3.898, 3.896, -1.95, 1.948]}