Answer to Question #3713 in Differential Equations for zirhyl kasek
Find the critical points and test for the maximun and minimum. y=x^3+3x^2+6x-4
y=x3 + 3x2 + 6x - 4
y'=0 3x2 + 6x + 6 = 0 x2 + 2x + 2 = 0 D = 4-4×1×2 < 0 There are no real solutions. Thus there is no minimum or maximum of this function on x∈R. Let’s find the second derivative : y'' = 2x+2 = 0 x = -1 is the inflection point.
If x < -1 then y'' < 0 graph is concave down If x > -1 then y'' > 0 graph is concave up.