The graph of f(x) = 2x^3 + 12 x^2 - 126 x + 16 has two horizontal tangents. One occurs at a negative value of x and the other at a positive value of x. What is the negative value of x where a horizontal tangent occurs?
What is the positive value of x where a horizontal tangent occurs?

Question

The graph of f(x) = 2x^3 + 12 x^2 - 126 x + 16 has two horizontal tangents. One occurs at a negative value of x and the other at a positive value of x. What is the negative value of x where a horizontal tangent occurs?
What is the positive value of x where a horizontal tangent occurs?

anonymous · Answer

$$f'(x) = 6x^2 + 24x -126$$
Horizontal tangent implies, slope must be Zero. Equating the slope to Zero, 
$$6x^2 + 24x - 126 = 0$$Now get the value of x, I hope you can do it from here.

anonymous · Answer

This is the same kind of problem we did before, but since it's a cubic equation, a little more complicated.  The process steps are the same.

anonymous · Answer

THANKS

anonymous · Answer

X=7 X=3