For the Function f(x)=x^3-3x+2, locate the ordered pairs of the absolute extrema on the interval [-1,3]

Answer 1

Take the derivative of this, set it equal to zero and solve. These will be the critical values and are possible extrema.

f'(x)=3x²-3

0=3(x²-1)

0=3(x-1)(x+1)

The CV are x=-1,1 and 3 (We also have to consider the end point of the interval that is not already a CV

We have to do a sign chart for this to determine if the extrema are mins or max. If the derivative changes from + to -, it is a relative max. If it changes from - to + it is a relative min.

-2 +

-1 CV relative max

0 -

1 CV relative min

2 +

Now, look at the y values at the CV and endpoint, use the original equation.

(-1,4)

(1,0) Absolute min

(3,20) Absolute max (20 is higher than 4 for the y values)