How do you find the local max and min for # y = 3x^4 + 4x^3 – 12x^2 + 1#?
Overall maximum is 1 at x = 0..
Local minimum
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To find the local maximum and minimum points of ( y = 3x^4 + 4x^3  12x^2 + 1 ), follow these steps:

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

Set the first derivative equal to zero and solve for ( x ) to find critical points. [ 12x^3 + 12x^2  24x = 0 ]

Factor out ( 12x ) from the equation. [ 12x(x^2 + x  2) = 0 ]

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 ]

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 ]

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

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

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 ]

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) ).
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