never, since the equation specifies that 6x^3-58x^2+130x+50>0
However, 6x^3-58x^2+130x+50 = 2(x-5)^2 (3x+1)
Since the leading coefficient is positive, we know that
as x→-∞, f(x)→-∞
Now, the graph crosses the x-axis at (-1/3,0) so
f(x) < 0 on (-∞,-1/3)
Since x=5 is a double root, the graph does not cross the x-axis there, meaning f(x) > 0 on (-1/3,5)U(5,∞)
determine when the function 6x^3-58x^2+130x+50>0 is less than 0
2 answers
So while the function is never less than 0, it is negative on the interval (-∞,-1/3) and positive on the intervals (-1/3,5)U(5,∞).