Use the derivative of f(x) = 6x7 + 7x6 to determine any points on the graph of f(x) at which the tangent line is horizontal.
f(x) = 6x^7 + 7x^6
f'(x) = 42x^6 + 42x^5
= 42x^5(x+1)
so, now it's easy to see where f'(x) = 0
f'(x) = 42x^6 + 42x^5
= 42x^5(x+1)
so, now it's easy to see where f'(x) = 0