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?

2 Answers
Aug 4, 2017

# (0,5) \ \ \ \ \ \ \ = # minimum
# (1,8) \ \ \ \ \ \ \ = # maximum
# (1/2,13/2) = # non-stationary inflection point

Explanation:

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

Aug 4, 2017

The local maximum is #=(1,8)# and the local minimum is #=(0,5)#
The point of inflection is #=(1/2,13/2)#

Explanation:

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]}