How do you find the critical points for # y = x^3 + 12x^2 + 6x + 8 #?

Question

Luke Alvarez · Accepted Answer

Since #lim_(x->+-oo) f(x)=+-oo# there is no global maximum or minimum. It has a local maximum at #x=-4-sqrt14# which is #f(-4-sqrt14)=28(4+sqrt14)# and it has local minimum at #x=-4+sqrt14# which is #f(-4+sqrt14)=-28(sqrt14-4)#

Scarlett Allison · Answer

undefined To find the critical points of $ y = x^3 + 12x^2 + 6x + 8 $, you need to first find the derivative of the function and then solve for points where the derivative equals zero.

The derivative of the function $ y $ with respect to $ x $ is $ y' = 3x^2 + 24x + 6 $.

Set $ y' $ equal to zero and solve for $ x $:

$ 3x^2 + 24x + 6 = 0 $

Now, you can use methods like factoring, completing the square, or the quadratic formula to solve for $ x $.

In this case, you can use the quadratic formula:

$ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} $

where $ a = 3 $, $ b = 24 $, and $ c = 6 $.

Substitute these values into the formula and solve for $ x $. After finding the values of $ x $, these are the critical points of the function.