What are the critical values of #f(x)=(x-lnx)/e^(x+ln(x^2))-x#?

1 Answer
Aug 19, 2017

Answer:

#f# has no critical values.

Explanation:

Note the simplification that can occur in the denominator: #e^(x+ln(x^2))=e^xe^ln(x^2)=x^2e^x#.

We also might note that the domain of #f# is #x>0#.

#f(x)=(x-lnx)/(x^2e^x)-x#

To find critical values, we find where the derivative #f'# equals #0# or does not exist within the domain of #f#.

To differentiate #f#, we'll need the quotient rule:

#f'(x)=((d/dx(x-lnx))x^2e^x-(x-lnx)(d/dxx^2e^x))/(x^2e^x)^2-1#

#f'(x)=((1-1/x)x^2e^x+(lnx-x)(2xe^x+x^2e^x))/(x^4e^(2x))-1#

Factoring #e^x# from the numerator and taking one #x# from #x^2# into the parentheses:

#f'(x)=(x(x-1)+(lnx-x)x(2+x))/(x^4e^x)-1#

#f'(x)=(x-1+(lnx-x)(x+2))/(x^3e^x)-1#

The derivative doesn't exist for #x<=0#, which is the same as #f#, so there are no critical points to be found there.

Other critical points could occur when #f'(x)=0#:

#0=(x-1+(lnx-x)(x+2))/(x^3e^x)-1#

Continuing to manipulate gives an equation which cannot be solved algebraically:

#x-1+(lnx-x)(x+2)-x^3e^x=0#

Graph this to find that is has no zeros.

This means that #f# has no critical values.