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

Aug 19, 2017

$f$ has no critical values.

#### Explanation:

Note the simplification that can occur in the denominator: ${e}^{x + \ln \left({x}^{2}\right)} = {e}^{x} {e}^{\ln} \left({x}^{2}\right) = {x}^{2} {e}^{x}$.

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

$f \left(x\right) = \frac{x - \ln x}{{x}^{2} {e}^{x}} - x$

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:

$f ' \left(x\right) = \frac{\left(\frac{d}{\mathrm{dx}} \left(x - \ln x\right)\right) {x}^{2} {e}^{x} - \left(x - \ln x\right) \left(\frac{d}{\mathrm{dx}} {x}^{2} {e}^{x}\right)}{{x}^{2} {e}^{x}} ^ 2 - 1$

$f ' \left(x\right) = \frac{\left(1 - \frac{1}{x}\right) {x}^{2} {e}^{x} + \left(\ln x - x\right) \left(2 x {e}^{x} + {x}^{2} {e}^{x}\right)}{{x}^{4} {e}^{2 x}} - 1$

Factoring ${e}^{x}$ from the numerator and taking one $x$ from ${x}^{2}$ into the parentheses:

$f ' \left(x\right) = \frac{x \left(x - 1\right) + \left(\ln x - x\right) x \left(2 + x\right)}{{x}^{4} {e}^{x}} - 1$

$f ' \left(x\right) = \frac{x - 1 + \left(\ln x - x\right) \left(x + 2\right)}{{x}^{3} {e}^{x}} - 1$

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

Other critical points could occur when $f ' \left(x\right) = 0$:

$0 = \frac{x - 1 + \left(\ln x - x\right) \left(x + 2\right)}{{x}^{3} {e}^{x}} - 1$

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

$x - 1 + \left(\ln x - x\right) \left(x + 2\right) - {x}^{3} {e}^{x} = 0$

Graph this to find that is has no zeros.

This means that $f$ has no critical values.