How do you find the properties of the function #f(x) = (x-1)ln(x-1)# (domain, intercepts, turning points, odd/even, etc.) ?

2 Answers
Jan 23, 2018

See graph as a guid

Explanation:

Tony B

Jan 24, 2018

See explanation...

Explanation:

Given:

#f(x) = (x-1)ln(x-1)#

Conventionally, the domain of #f(x)# is the set of real values of #x# such that #f(x)# is defined.

Provided #x > 1#, the logarithm will be defined and hence #f(x)# will be too.

So the (implicit) domain of #f(x)# is #(1, oo)#

Note that #f(x)# is not defined for #x in (-oo, -1)#, so #f(x)# is neither an even nor odd function. Neither can it be periodic.

The only #x# intercept is when #ln(x-1) = 0#, which is when #x=2#, since the other factor #(x-1)# is non-zero for all #x in (1, oo)#.

When #x in (1, 2)# we have #(x-1) > 0# and #ln(x-1) < 0#, so #f(x) < 0#

When #x in (2, oo)# we have #(x-1) > 0# and #ln(x-1) > 0#, so #f(x) > 0#.

#f'(x) = (d/(dx) (x-1))ln(x-1) + (x-1)(d/(dx) ln(x-1))#

#color(white)(f'(x)) = ln(x-1)+(x-1)/(x-1)#

#color(white)(f'(x)) = ln(x-1)+1#

So #f'(x) = 0# when #ln(x-1) = -1#, i.e. when #x=1+1/e#

If #x in (1, 1+1/e)# then #f'(x) < 0# so #f(x)# is strictly monotonically decreasing.

If #x in (1+1/e, oo)# then #f'(x) > 0# so #f(x)# is strictly monotonically increasing.

So #f(x)# has a local minimum - which is also a global minimum - when #x=1+1/e# and #f(1+1/e) = -1/e#

#f^((2)) = 1/(x-1) > 0# for all #x in (1, oo)#

So #f(x)# has no point of inflection.