Potential solutions are the function values at the boundary of the interval and any relative extrema that occur within the interval.

Boundary Values:

f(-1) = -4 - 8 + 1 = -11

f(1) = 4 - 8 + 1 = -3

Relative Extrema:

f'(x) = 0 = 12x² - 16x = 4x(3x-4)

x = 0, 4/3

f(0) = 1

x = 4/3 is outside the interval and can be ignored

The set of potential solutions is {-11, -3, 1}

Ans: Minimum value on the interval is -11 at x =-1

f'(x) = 12x^2 - 16x = 0

4x(3x - 4) = 0.

Critical numbers are at 0 and (4/3).

f''(x) = 24x - 16

At x = 0, the second derivative is -16 and at x = 4/3 the second derivative 16.

So, the value of the function at x = 4/3 is concave up and therefore is a local minimum.

f(4/3) = 4(4/3)^3 - 8(4/3)^2 + 1 = -101/27

f(-1) = -11

f(1) = -3

The absolute minimum occurs at x = -1 and is -11.

f ' = 12 x² - 16 x .....= 0 when x = 0 or 4/3....thus just consider f (-1) vs f(0) vs f(1)