How do you find the local max and min for # y = 3x^4 + 4x^3 – 12x^2 + 1#?

Question

Charles Anctil · Accepted Answer

Overall maximum is 1 at x = 0..
Local minimum #-4# at x =1 and #-31# at x = #-2#.

#dy/dx = 12(x^3+x^2-2 x) = 12 x (x^2+x-2)=12x(x-1)(x+2)#. = 0, at x 0, x=1 and x=#-2#. The second derivative is #12 (3x^2+2 x-2)# This is < 0 at x = 0 and > 0 at x = 1 and x = #-2#. Accordingly, y is the maximum at x = 0 and local minimum, at each of x - 1 and x =#-2#.

Luke Alvarez · Answer

undefined To find the local maximum and minimum points of $ y = 3x^4 + 4x^3 - 12x^2 + 1 $, follow these steps:

1. Compute the first derivative of the function to find critical points.
   $$ y' = 12x^3 + 12x^2 - 24x $$

2. Set the first derivative equal to zero and solve for $ x $ to find critical points.
   $$ 12x^3 + 12x^2 - 24x = 0 $$

3. Factor out $ 12x $ from the equation.
   $$ 12x(x^2 + x - 2) = 0 $$

4. Solve for $ x $ using the quadratic equation $ x^2 + x - 2 = 0 $.
   $$ x^2 + x - 2 = 0 $$
   $$ (x + 2)(x - 1) = 0 $$
   $$ x = -2, \, x = 1 $$

5. Plug the critical points into the original function to find the corresponding $ y $ values.
   $$ y(-2) = 3(-2)^4 + 4(-2)^3 - 12(-2)^2 + 1 = 49 $$
   $$ y(1) = 3(1)^4 + 4(1)^3 - 12(1)^2 + 1 = -4 $$

6. The critical points $ (-2, 49) $ and $ (1, -4) $ represent potential local maximum and minimum points.

7. To confirm whether these points are local maxima or minima, use the second derivative test.
   $$ y'' = 36x^2 + 24x - 24 $$

8. Plug the critical points into the second derivative.
   $$ y''(-2) = 36(-2)^2 + 24(-2) - 24 = -48 $$
   $$ y''(1) = 36(1)^2 + 24(1) - 24 = 36 $$

9. Since $ y''(-2) < 0 $, the point $ (-2, 49) $ is a local maximum.
   Since $ y''(1) > 0 $, the point $ (1, -4) $ is a local minimum.

Therefore, the local maximum point is $ (-2, 49) $ and the local minimum point is $ (1, -4) $.