How do you find all relative extrema and points of inflection for the equation: #y=x^2 log_3x#?

1 Answer
Jul 5, 2015

Answer:

Find extrema using the derivative and points of inflection using the second derivative.

Explanation:

We will make use of: #log_3x = lnx/ln3# and

#d/dx log_3x = d/dx(lnx/ln3) = 1/(xln3)#

Finding relative extrema

#f(x) = y=x^2 log_3x#

Note that the domain of this function is #(0,oo)#

Use the product rule to get:

#y' = 2xlog_3x + x^2 * 1/(xln3)#

# = 2xlog_3x + x/ln3#

To find critical numbers, it is convenient to use all logarithms with the same base.:

#f'(x) = y' = (2xlnx)/ln3 + x/ln3 = (x(2lnx+1))/ln3#

#y'# in never undefined in the domain, and is equal to #0# at

#x=0# or #2lnx+1=0#. #0# is not in the domain, so we need only solve:
#2lnx+1=0#.
#lnx = -1/2#

#x = e^(-1/2) = 1/sqrte#

The partitions on the domain are: #(0, e^(-1/2)), (e^(-1/2), oo)#

Testing #(0, e^(-1/2))#.
Note that #1/e < 1/ sqrte#, so #e^-1# is in the first interval and

#f'(e^(-1)) = (e^(-1)(2lne^(-1) +1))/ln3 = (1/e(-2+1))/ln3# which is negative.

So #f'# is negative on #(0, e^(-1/2))#.

Testing #(e^(-1/2), oo)#
Use #1# as the test number.

#f'(e) = (1(2ln1+1))/ln3 = (1 (0+1))/ln3# which is positive, so

So #f'# is positive on #(e^(-1/2), oo)#.

This tells us that #f(e^(-1/2)) is a relative minimum.

Finding points of inflection

Recall:

#f'(x) = y' = # = 2xlog_3x + x/ln3#

So

#f''(x) = y'' = 2log_3x +2x * 1/(xln3) +1/ln3#

# = (2lnx)/ln3 +2/ln3 + 1/ln3 = (2lnx + 3)/ln3#

The only partition number for #f''# is #lnx = -3/2# so #x = e^(-3/2)#

For #f''# the parition intervals are #(0, e^(-3/2)), (e^(-3/2), oo)#

Using #e^-2# and #1# as test numbers we will find that:

#f''# is negative on #(0, e^(-3/2))#.

#f'# is positive on #(e^(-3/2), oo)#.

The concavity does change, so

#(e^(-3/2), f(e^(-3/2)))# is a point of inflection.