How do you find the critical points for #y = 2x^3 + 3x^2 - 12x + 4#?

Question

Charles Arocha · Accepted Answer

The critical points are #x=-2# and #x=1#. You evaluate the first derivative and find its zeroes: #y'=2*3x^2+3*2x-12*1+0=6x^2+6x-12#
#y'=6x^2+6x-12#
And now we can use the quadratic formula
#Delta=6^2-4*6*(-12)=324#
#sqrt(Delta)=18#
#x_1=(-6-18)/(2*6)=(-24)/12=-2#
#x_2=(-6+18)/(2*6)=12/12=1#

Adam Adkins · Answer

undefined To find the critical points of the function y = 2x^3 + 3x^2 - 12x + 4, you first need to find its derivative, y'. Then, set y' equal to zero and solve for x. The values of x obtained from this equation will be the critical points.

y' = 6x^2 + 6x - 12

Set y' equal to zero:
6x^2 + 6x - 12 = 0

Now, solve for x using methods like factoring, completing the square, or the quadratic formula.

Using the quadratic formula:
x = [-b ± √(b^2 - 4ac)] / (2a)

Where a = 6, b = 6, and c = -12.

Plugging in these values:
x = [-6 ± √(6^2 - 4 * 6 * (-12))] / (2 * 6)

x = [-6 ± √(36 + 288)] / 12

x = [-6 ± √324] / 12

x = [-6 ± 18] / 12

So, the critical points are:
x = (-6 + 18) / 12 = 12 / 12 = 1
x = (-6 - 18) / 12 = -24 / 12 = -2

Therefore, the critical points for the function y = 2x^3 + 3x^2 - 12x + 4 are x = 1 and x = -2.