What are the absolute extrema of f ( x ) = x 3 - 3 x + 1 ∈ [ 0 , 3 ] ?

Question

What are the absolute extrema of             f              (        x        )              =          x      3        -    3    x    +    1    ∈          [      0      ,      3      ]      ?

Cheyenne Lynn · Accepted Answer

There are two potential candidates for an interval's absolute extrema. They are the endpoints of the interval (here,

0

and

3

) and the critical values of the function located within the interval.
The critical values can be found by finding the function's derivative and finding for which values of

x

it equals

0

.
We can use the power rule to find that the derivative of

f (x) = x^{3} - 3 x + 1

is

f' (x) = 3 x^{2} - 3

.
The critical values are when

3 x^{2} - 3 = 0

, which simplifies to be

x = \pm 1

. However,

x = - 1

is not in the interval so the only valid critical value here is the one at

x = 1

. We now know that the absolute extrema could occur at

x = 0, x = 1,

and

x = 3

.
To determine which is which, plug them all into the original function.

f (0) = 1

f (1) = - 1

f (3) = 19

From here we can see that there is an absolute minimum of

- 1

at

x = 1

and an absolute maximum of

19

at

x = 3

.
Check the function's graph:
graph{x^3-3x+1 [-0.1, 3.1, -5, 20]}

What are the absolute extrema of f(x)= x^3 -3x+1 in [0,3]?

Answered question

Answer & Explanation

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