How do you find the local maximum and minimum values of #f(x) = 2x^3 - 5x +1# in the the interval is (-3,3)?

Question

How do you find the local maximum and minimum values of #f(x) = 2x^3 - 5x +1# in the  the interval is (-3,3)?

Christopher Agresta · Accepted Answer

#x=-sqrt(5/6)# is a local maximum
#x=sqrt(5/6)# is a local minimum
graph{2x^3-5x+1 [-10, 10, -5, 5]}

Find local extrema on the interval by finding where #f'(x)# is equal to zero. First find #f'(x)# #f(x)=2x^3-5x+1# #f'(x)=6x^2-5# #0=6x^2-5# #5=6x^2# #x^2=5/6# #x=+-sqrt(5/6)# #xapprox+-.9129# Find whether each is a local max or local min by checking values around #+-sqrt(5/6)#; #x=-sqrt(5/6)# is a local maximum #x=sqrt(5/6)# is a local minimum

Emma Aldridge · Answer

undefined To find the local maximum and minimum values of $ f(x) = 2x^3 - 5x + 1 $ in the interval $(-3, 3)$, follow these steps:

1. Find the critical points by setting the derivative of $ f(x) $ equal to zero and solving for $ x $.
   
   $ f'(x) = 6x^2 - 5 $
   
   Set $ f'(x) = 0 $ and solve for $ x $:
   
   $ 6x^2 - 5 = 0 $
   
   $ 6x^2 = 5 $
   
   $ x^2 = \frac{5}{6} $
   
   $ x = \pm \sqrt{\frac{5}{6}} $
   
   The critical points are $ x = -\sqrt{\frac{5}{6}} $ and $ x = \sqrt{\frac{5}{6}} $.

2. Evaluate $ f(x) $ at the critical points and at the endpoints of the interval $(-3, 3)$.
   
   $ f(-\sqrt{\frac{5}{6}}) \approx 4.63 $
   
   $ f(\sqrt{\frac{5}{6}}) \approx -2.63 $
   
   $ f(-3) = -47 $
   
   $ f(3) = 52 $

3. Compare the values obtained in step 2 to determine which one is the maximum and which one is the minimum within the given interval.

The local maximum value is $ f(3) = 52 $.
   
   The local minimum value is $ f(-\sqrt{\frac{5}{6}}) \approx 4.63 $.