# What are the critical values, if any, of f(x)= 5x + 6x ln x^2?

Feb 5, 2018

$x = {e}^{- \frac{17}{12}} \approx 0.659241 \setminus \setminus \left(6 \setminus \mathrm{dp}\right)$

#### Explanation:

We have:

$f \left(x\right) = 5 x + 6 x \ln \left({x}^{2}\right)$

Which, using the properties of logarithms, we can write as:

$f \left(x\right) = 5 x + 6 x \left(2\right) \ln \left(x\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = 5 x + 12 x \ln \left(x\right)$

Then, differentiating wrt $x$ by applying the product rule we get:

$f ' \left(x\right) = 5 + \left(12 x\right) \left(\frac{1}{x}\right) + \left(12\right) \left(\ln x\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 5 + 12 + 12 \ln x$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 17 + 12 \ln x$

At a critical point, we requite that the first derivative vanishes, thus we require that:

$f ' \left(x\right) = 0$

$\therefore 17 + 12 \ln x = 0$
$\therefore \ln x = - \frac{17}{12}$
$\therefore x = {e}^{- \frac{17}{12}} \approx 0.659241 \setminus \setminus \left(6 \setminus \mathrm{dp}\right)$