# Answer to Question #3713 in Differential Equations for zirhyl kasek

Question #3713

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

Expert's answer

y=x

y'=0

3x

x

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.

^{3}+ 3x^{2}+ 6x - 4y'=0

3x

^{2}+ 6x + 6 = 0x

^{2}+ 2x + 2 = 0D = 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.

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## Comments

Assignment Expert01.08.11, 13:28D is the discrimminant of the square equation.

zirhyl kasek30.07.11, 05:52Thank you for answering my question. That was just what I thought but wasn't quite sure. What do you mean by D= 4-4*1*2?

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