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

##### 1 Answer

#### Explanation:

Note the simplification that can occur in the denominator:

We also might note that the domain of

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

To find critical values, we find where the derivative

To differentiate

#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

#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

Other critical points could occur when

#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