What are the extrema of #f(x) = x^3 - 27x#?

Question

Paisley Johnson · Accepted Answer

#(-3, 54) and (3, -54)#

Relative maximum and minimum points occur when the derivative is zero, that is when #f'(x)=0#. So in this case, when #3x^2-27=0# #=>x=+-3.

Since the second derivative #f''(-3)<0 and f''(3)>0#, it implies that a relative maximum occurs at x=-3 and a relative minimum at x=3.

But: #f(-3)=54 and f(3)=-54# which implies that #(-3, 54)# is a relative maximum and #(3, -54)# is a relative minimum.

graph{x^3-27x [-115.9, 121.4, -58.1, 60.5]}

Christopher Allers · Answer

undefined To find the extrema of $ f(x) = x^3 - 27x $, we first need to find its critical points by setting its derivative equal to zero and solving for $ x $.

1. Find the derivative of $ f(x) $:
$$ f'(x) = 3x^2 - 27 $$

2. Set the derivative equal to zero and solve for $ x $:
$$ 3x^2 - 27 = 0 $$
$$ 3x^2 = 27 $$
$$ x^2 = 9 $$
$$ x = \pm 3 $$

3. Now, we need to test these critical points to determine whether they correspond to maximum, minimum, or neither.

- $ x = -3 $:
  Substitute $ x = -3 $ into the second derivative:
  $ f''(-3) = 6(-3) = -18 $
  Since the second derivative is negative, $ x = -3 $ corresponds to a local maximum.

- $ x = 3 $:
  Substitute $ x = 3 $ into the second derivative:
  $ f''(3) = 6(3) = 18 $
  Since the second derivative is positive, $ x = 3 $ corresponds to a local minimum.

So, the local maximum occurs at $ x = -3 $ and the local minimum occurs at $ x = 3 $. These are the extrema of the function $ f(x) = x^3 - 27x $.