Question

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

Answer 1

`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.