How do you find the local extrema for #f(x)=-0.12x^3 + 900x - 830#?

Question

Elijah Abram · Accepted Answer

local max. at #(50, 29170)#
local min. at #(-50, -30830)# #f(x) = -.12x^3 + 900x -830# Find the first derivative: #f'(x) = -.36x^2+900# Find the critical numbers (#f'(x) = 0#): #.36x^2 = 900; x^2 = 2500; x = +-50# Second derivative test: #f''(x) = -.72x;# #f''(-50) > 0#; relative min at #x = -50# and #f(-50) = -30,830# #f''(50) < 0#; relative max at #x = 50# and #f(50) = 29,170# local max. at #(50, 29170)# local min. at #(-50, -30830)#

Samantha Adkison · Answer

undefined To find the local extrema of $ f(x) = -0.12x^3 + 900x - 830 $, you first need to find its critical points by taking the derivative of the function and setting it equal to zero. Then, you can determine whether each critical point corresponds to a local minimum, local maximum, or neither by using the second derivative test or by examining the behavior of the function around each critical point.

1. Find the derivative of the function:
$$ f'(x) = -0.36x^2 + 900 $$

2. Set the derivative equal to zero and solve for $ x $ to find the critical points:
$$ -0.36x^2 + 900 = 0 $$
$$ -0.36x^2 = -900 $$
$$ x^2 = \frac{900}{0.36} $$
$$ x^2 = 2500 $$
$$ x = \pm 50 $$

3. Test each critical point to determine the nature of the extremum:

- Test $ x = 50 $:
     - Calculate the second derivative: $ f''(x) = -0.72x $
     - $ f''(50) = -0.72(50) = -36 < 0 $
     - Since the second derivative is negative, $ x = 50 $ corresponds to a local maximum.
   
   - Test $ x = -50 $:
     - Calculate the second derivative: $ f''(x) = -0.72x $
     - $ f''(-50) = -0.72(-50) = 36 > 0 $
     - Since the second derivative is positive, $ x = -50 $ corresponds to a local minimum.

Therefore, the local maximum occurs at $ x = 50 $ and the local minimum occurs at $ x = -50 $.