How do you find the local maximum and minimum values of #f(x) = 5 + 9x^2 − 6x^3# using both the First and Second Derivative Tests?

Question

Scarlett Allison · Accepted Answer

The local maximum is #=(1,8)# and the local minimum is #=(0,5)#
The point of inflection is #=(1/2,13/2)# Our function is #f(x)=5+9x^2-6x^3# The first derivative is #f'(x)=18x-18x^2# The critical points are when #f'(x)=0# #18x-18x^2=18x(1-x)# #18x(1-x)=0# Therefore, #x=0# and #x=1# We can build a variation chart #color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##0##color(white)(aaaaaaa)##1##color(white)(aaaa)##+oo# #color(white)(aaaa)##x##color(white)(aaaaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+# #color(white)(aaaa)##1-x##color(white)(aaaaa)##+##color(white)(aaaa)##+##color(white)(aaaa)##-# #color(white)(aaaa)##f'(x)##color(white)(aaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##-# #color(white)(aaaa)##f(x)##color(white)(aaaaa)##↘##color(white)(aaaa)##↗##color(white)(aaaa)##↘# Now, we calculate the second derivative #f''(x)=18-36x# The point of inflection is when #f''(x)=0# That is, #18-36x=0#, #=>#, #x=18/36=1/2# We build a variation chart with the second derivative #color(white)(aaaa)##Interval##color(white)(aaaa)##(-oo,1/2)##color(white)(aaaa)##(1/2,+oo)# #color(white)(aaaa)##sign f''(x)##color(white)(aaaaaa)##+##color(white)(aaaaaaaaaaaa)##-# #color(white)(aaaa)## f(x)##color(white)(aaaaaaaaaaaa)##uu##color(white)(aaaaaaaaaaaa)##nn# graph{5+9x^2-6x^3 [-15.35, 16.69, -3.14, 12.88]}

Paisley Johnson · Answer

# (0,5) \ \ \ \ \ \ \ = # minimum
# (1,8) \ \ \ \ \ \ \ = # maximum
# (1/2,13/2) = # non-stationary inflection point We have: # f(x) = 5 + 9x^2-6x^3 # We can see the critical point via a graph: graph{5 + 9x^2-6x^3 [-6, 6, -2, 14]} We can examine the critical points using calculus: Differentiating wrt #x# we get: # f'(x) = 18x - 18x^2 # At a critical point we have #f'(x) = 0# # f'(x) = 0 => 18x - 18x^2 = 0 # # :. 18x(1-x) = 0 => x = 0,1 # And, now we have the #x#-coordinates, we can determine the nature of the turning points (or critical points) by using the second derivative test. Differentiating a second time, we get: # f''(x) = 18 - 36x # When: # x= 0 => f''(0) = 18-0 \ \ gt 0 =>#minimum # x= 1 => f''(1) = 18-36 lt 0 => #maximum Also note we have an inflection point if #f''(x)=0# # f''(x) = 0 => 18-36x = 0 => x=1/2 # Now we have the #x#-coordinate of the critical points let us find the associated #y#-coordinate: # x= 0 \ => f(0) \ \ \ \ \ = 5 + 0-0 = 5 # # x= 1 \ => f(1) \ \ \ \ \ = 5 + 9-6 = 8 # # x= 1/2 => f(1/2) = 5 + 9(1/4)-6(1/8) = 13/2 # Hence, in summary # (0,5) \ \ \ \ \ \ \ = # minimum # (1,8) \ \ \ \ \ \ \ = # maximum # (1/2,13/2) = # non-stationary inflection point Which is consistent with what we see graphically

Christopher Agresta · Answer

undefined To find the local maximum and minimum values of $ f(x) = 5 + 9x^2 - 6x^3 $ using both the First and Second Derivative Tests:

1. **First Derivative Test**:
   - Find the critical points by setting the derivative equal to zero and solving for $ x $.
   - Take the first derivative of $ f(x) $ and set it equal to zero:
     $$ f'(x) = 18x - 18x^2 $$
   - Set $ f'(x) $ equal to zero and solve for $ x $:
     $$ 18x - 18x^2 = 0 $$
     $$ 18x(1 - x) = 0 $$
     $$ x = 0 \text{ or } x = 1 $$
   - These are the critical points.

2. **Second Derivative Test**:
   - Find the second derivative of $ f(x) $ and evaluate it at the critical points.
     $$ f''(x) = 18 - 36x $$
   - Evaluate $ f''(x) $ at the critical points:
     - For $ x = 0 $: $ f''(0) = 18 > 0 $ (Concave up)
     - For $ x = 1 $: $ f''(1) = 18 - 36(1) = -18 < 0 $ (Concave down)

3. **Conclusion**:
   - At $ x = 0 $, since the second derivative is positive, it implies a local minimum.
   - At $ x = 1 $, since the second derivative is negative, it implies a local maximum.

Therefore, $ f(x) $ has a local minimum at $ x = 0 $ and a local maximum at $ x = 1 $.