We have, f(x) = 3x4 + 4x3 - 12x2 + 5
∴ f'(x) = 12x3 + 12x2 - 24x
= 12x (x2+x-2)
= 12x (x2+2x-x-2)
= 12x (x-1)(x+2)
Therefore, f'(x) = 0 gives x = 0 or x = 1 or x = -2.
Hence, critical points for given function are x = 0, x = -2 and x = 1.
Now, f''(x) = 36x2 + 24x - 24
∵ f''(0) = -24 < 0
∴ x = 0 is point of local maxima (By second derivative test)
Also f''(1) = 36+24-24 = 36 > 0
∴ x = 1 is point of local minima (By second derivative test)
Now, f''(-2) = 36 x 4 + 24 x -2 - 24
= 144 - 48 - 24
= 144 - 72
= 72 > 0
∴ x = -2 is point of local minima (By second derivative test)
Now, f(0) = 5, f(1) = 3+4-12+5 = 12 - 12 = 0
f(-2) = 48 - 32 - 48 + 5 = -32+5 = -27
Hence, (1, 0) and (-2, -27) are point of local minima and (0, 5) is point of local maxima.