How do you locate the absolute extrema of the function #f(x)=x^3-12x# on the closed interval [0,4]?

Question

Christopher Allers · Accepted Answer

The absolute extrema of a function, #f#, that is continuous on a closed interval must occur at either a critical number for #f# in the interval, or at an endpoint of the interval.

So consider: #f(x) =x^3-12x# #f'(x) = 3x^2-12#. #f'# is polynomial, so every critical number for #f# is a zero for #f'#. The zeros of #f'# are: #3(x^2-4)=0#, #x=-2, 2#. But #-2# is not in the interval #[0,4]#. So the only critical number in the interval is #2#. The minimum and maximum must occur at one of the values #x=# #0# or #2# or #4#. To finish, evaluate #f# at each of these values. #f(0) = 0# #f(2) = 8-24 = -16# #f(4) = 64-48 = 16# The minimum is #-16# (it occurs at #2#) The maximum is #16# (it occurs at #4#). To understand what we've done it may help to see the graph: graph{y=(x^3-12x)*sqrt(4-(x-2)^2)/(sqrt(4-(x-2)^2)) [-52.2, 40.28, -23.87, 22.4]}

Luke Alvarez · Answer

undefined To locate the absolute extrema of the function $ f(x) = x^3 - 12x $ on the closed interval $[0,4]$, follow these steps:

1. Find the critical points of the function by setting the derivative equal to zero and solving for $ x $.
2. Evaluate the function at the critical points and at the endpoints of the interval.
3. The maximum and minimum values among these critical points and endpoints will be the absolute extrema.

Let's proceed with these steps:

1. Find the derivative of the function: 
   $ f'(x) = 3x^2 - 12 $.

2. Set the derivative equal to zero and solve for $ x $:
   $ 3x^2 - 12 = 0 $.
   Solving this equation gives $ x = \pm 2 $.

3. Evaluate the function at these critical points and at the endpoints:
   $ f(0) = 0 $,
   $ f(2) = 2^3 - 12(2) = -16 $,
   $ f(4) = 4^3 - 12(4) = 16 - 48 = -32 $.

4. Compare the values obtained:
   - $ f(0) = 0 $
   - $ f(2) = -16 $
   - $ f(4) = -32 $

The absolute maximum is $ f(0) = 0 $, and the absolute minimum is $ f(4) = -32 $.

Emilia Aguilar · Answer

undefined To locate the absolute extrema of the function $ f(x) = x^3 - 12x $ on the closed interval $[0,4]$, you follow these steps:

1. Find the critical points of the function within the interval by setting the derivative equal to zero and solving for $x$.
2. Evaluate the function at the critical points and the endpoints of the interval.
3. Identify the highest and lowest values among these values to determine the absolute maximum and minimum.

Let's proceed with the steps:

1. Find the derivative of the function:
$$ f'(x) = 3x^2 - 12 $$
Set $ f'(x) $ equal to zero and solve for $ x $ to find critical points:
$$ 3x^2 - 12 = 0 $$
$$ x^2 - 4 = 0 $$
$$ (x - 2)(x + 2) = 0 $$
$$ x = 2 \text{ or } x = -2 $$

2. Evaluate the function $ f(x) $ at the critical points and the endpoints:
$$ f(0) = (0)^3 - 12(0) = 0 $$
$$ f(2) = (2)^3 - 12(2) = -16 $$
$$ f(4) = (4)^3 - 12(4) = -32 $$

3. Compare these values to find the absolute maximum and minimum:
The values are: $ f(0) = 0 $, $ f(2) = -16 $, and $ f(4) = -32 $.
The absolute maximum is $ f(0) = 0 $ and the absolute minimum is $ f(4) = -32 $.

So, the absolute maximum occurs at $ x = 0 $ and the absolute minimum occurs at $ x = 4 $.