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

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Answer 1 · 2018-02-05T12:39:28

$x = e^(-17/12) ~~ 0.659241 \ \ (6 \ dp)$

We have:

$f(x) = 5x+6xln(x^2)$

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

$f(x) = 5x+6x(2)ln(x)$
$\ \ \ \ \ \ \ = 5x+12xln(x)$

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

$f'(x) = 5 + (12x)(1/x) + (12)(lnx)$
$\ \ \ \ \ \ \ \ \ = 5+12+12lnx$
$\ \ \ \ \ \ \ \ \ = 17+12lnx$

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

$f'(x) = 0$

$:. 17+12lnx = 0$
$:. lnx = -17/12$
$:. x = e^(-17/12) ~~ 0.659241 \ \ (6 \ dp)$

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