We will make use of: log_3x = lnx/ln3log3x=lnxln3 and
d/dx log_3x = d/dx(lnx/ln3) = 1/(xln3)ddxlog3x=ddx(lnxln3)=1xln3
Finding relative extrema
f(x) = y=x^2 log_3xf(x)=y=x2log3x
Note that the domain of this function is (0,oo)(0,∞)
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.
