How do you use the second derivative test to find all relative extrema of #f(x)=5+3x^2-x^3#?

Question

Charles Arocha · Accepted Answer

The first derivative is #f'(x)=6x-3x^2=3x(2-x)#, which has roots at #x=0# and #x=2#. These are the critical point, and also the possible locations of local extrema. Since the second derivative is #f''(x)=6-6x#, we get #f''(0)=6>0# and #f''(2)=-6<0#. The fact that #f''(0)>0# (and the fact that #f''# is continuous) implies that the graph of #f# is concave up near #x=0#, making, by the Second Derivative Test, #x=0# the location of a local minimum. The fact that #f''(2)<0# (and the fact that #f''# is continuous) implies that the graph of #f# is concave down near #x=2#, making, by the Second Derivative Test, #x=2# the location of a local maximum. The local minimum value (output) is #f(0)=5# and the local maximum value (output) is #f(2)=5+12-8=9#.

Charles Arocha · Answer

undefined To use the second derivative test to find all relative extrema of $ f(x) = 5 + 3x^2 - x^3 $:

1. Find the first derivative of $ f(x) $, denoted by $ f'(x) $.
   $ f'(x) = 6x - 3x^2 $.

2. Find the second derivative of $ f(x) $, denoted by $ f''(x) $.
   $ f''(x) = 6 - 6x $.

3. Find the critical points by setting $ f'(x) = 0 $ and solving for $ x $.
   $ 6x - 3x^2 = 0 $
   $ 3x(2 - x) = 0 $
   $ x = 0 $ or $ x = 2 $.

4. Evaluate the sign of $ f''(x) $ at each critical point.
   $ f''(0) = 6 > 0 $ (Concave up, so it's a local minimum)
   $ f''(2) = -6 < 0 $ (Concave down, so it's a local maximum)

5. Thus, $ f(x) $ has a relative minimum at $ x = 0 $ and a relative maximum at $ x = 2 $.