use the derivative of f(x)=2x^3 + 6x to determine any points on the graph of f(x) at which the tangent line is horizontal
a. (1,8) and (-1,-8)
b. f(x) has no points with a horizontal tangent line
c. (2,28) and (-2,28)
d. (1,8)
e. (0,0)
when the tangent line is horizontal, the slope must be zero.
But the slope is equal to the derivative, so ....
f(x) = 2x^3 + 6x
f ' (x) = 6x^2 + 6
then 6x^2 + 6 = 0
6x^2 = -6
x^2 = -1 , good luck on that
Since our equation has not real solution, there is no point at which the tangent is horizontal.
check by looking at the graph using Wolfram:
http://www.wolframalpha.com/input/?i=plot+y+%3D+2x%5E3+%2B+6x