Given `f(x)=x^3-3x^2+1` on the interval `[-1/2,4]` , we are asked to find the absolute maximum and minimum.

Since the function is continuous, we are guaranteed a maximum and a minimum on the interval.

The extrema can only occur at critical points, or the endpoints of the interval. The critical points...

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Given `f(x)=x^3-3x^2+1` on the interval `[-1/2,4]` , we are asked to find the absolute maximum and minimum.

Since the function is continuous, we are guaranteed a maximum and a minimum on the interval.

The extrema can only occur at critical points, or the endpoints of the interval. The critical points are where the first derivative is zero or fails to exist. Since this function is a polynomial, it is differentiable so we only need the points where the derivative is zero.

`f'(x)=3x^2-6x`

Setting the derivative equal to zero we get:

`3x^2-6x=0`

`3x(x-2)=0`

x=0 or x=2 both of which occur in the interval.

We now evaluate f(x) at `-1/2,0,2,4` to determine the maximum and minimum:

`f(-1/2)=1/8`

`f(0)=1`

`f(2)=-3`

`f(4)=17`

**So the maximum occurs at x=4 and has a value of 17.****The minimum occurs at x=2 with as value of -3.**

The graph: