Question

Find the critical points and test for the maximun and minimum. y=x^3+3x^2+6x-4

Answer 1

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.