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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=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.
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.
Learn more about our help with Assignments: Differential Equations
Comments
D is the discrimminant of the square equation.
Thank 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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