# What are the critical points of  f(x) = e^xlnx^2?

Jul 18, 2017

The function $f \left(x\right) = {e}^{x} \ln \left({x}^{2}\right)$ has a single critical point in $\left(- \infty , 0\right)$ whose value is the solution of the equation:

$x \ln \left(- x\right) + 1 = 0$

#### Explanation:

Simplify the expression noting that based on properties of logarithms:

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

the critical points are by definition the points where $f ' \left(x\right) = 0$ so evaluate the derivative of the function:

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left(2 {e}^{x} \ln \left\mid x \right\mid\right) = 2 \left(\left(\frac{d}{\mathrm{dx}} {e}^{x}\right) \ln \left\mid x \right\mid + {e}^{x} \left(\frac{d}{\mathrm{dx}} \ln \left\mid x \right\mid\right)\right)$

$\frac{\mathrm{df}}{\mathrm{dx}} = 2 \left({e}^{x} \ln \left\mid x \right\mid + {e}^{x} / x\right)$

$\frac{\mathrm{df}}{\mathrm{dx}} = 2 {e}^{x} \left(\ln \left\mid x \right\mid + \frac{1}{x}\right)$

To identify the critical points then we have to solve the equation:

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

and as $2 {e}^{x} \ne 0$:

$\ln \left\mid x \right\mid = - \frac{1}{x}$

This is a logarithmic equation that can be solved only approximately:
note that for $x > 0$ we have $\ln x = - \frac{1}{x} < 0$ so the solution should be in the interval $\left(0 , 1\right)$.

However given:

$\ln x + \frac{1}{x} = 0$

as $x \ne 0$ this is equivalent to:

$x \ln x + 1 = 0$

Now consider the function $g \left(x\right) = x \ln x + 1$:

${\lim}_{x \to 0} \left(x \ln x + 1\right) = 1 + {\lim}_{x \to 0} x \ln x$

${\lim}_{x \to 0} \left(x \ln x + 1\right) = 1 + {\lim}_{x \to 0} \ln \frac{x}{\frac{1}{x}}$

Apply l'Hospital's:

${\lim}_{x \to 0} \left(x \ln x + 1\right) = 1 + {\lim}_{x \to 0} \frac{\frac{d}{\mathrm{dx}} \ln x}{\frac{d}{\mathrm{dx}} \frac{1}{x}}$

${\lim}_{x \to 0} \left(x \ln x + 1\right) = 1 + {\lim}_{x \to 0} \frac{\frac{1}{x}}{- \frac{1}{x} ^ 2} = 1 + {\lim}_{x \to 0} \left(- x\right) = 1$

while:

${\lim}_{x \to 1} \left(x \ln x + 1\right) = 1$

Consider now:

$\frac{d}{\mathrm{dx}} \left(x \ln x + 1\right) = 1 + \ln x$

so the only critical point for this function is in $x = \frac{1}{e}$ where its value is:

$g \left(\frac{1}{e}\right) = \frac{1}{e} \ln \left(\frac{1}{e}\right) + 1 = 1 - \frac{1}{e} > 0$

and as $g ' ' \left(\frac{1}{e}\right) = e > 0$ this is a local minimum. As the minimum value of the function is positive, this means that it is never zero. So actually we have no critical points for $x > 0$

On the other hand for $x < 0$ the equation becomes:

$x \ln \left(- x\right) + 1 = 0$

and substituting $t = - x$ we can see that:

${\lim}_{x \to {0}^{-}} \left(x \ln \left(- x\right) + 1\right) = 1 - {\lim}_{t \to {0}^{+}} t \ln t = 1$

while:

${\lim}_{x \to - \infty} \left(x \ln \left(- x\right) + 1\right) = - \infty$

thus the function assumes positive and negative values and as it is continuous in the interval $\left(- \infty , 0\right)$ it must vanish at least in one point.

graph{e^x ln(x^2) [-10, 10, -5, 5]}