What are the local extrema, if any, of f (x) =2ln(x^2+3)-x?

Jan 11, 2017

$f \left(x\right) = 2 \ln \left({x}^{2} + 3\right) - x$ has a local minimum for $x = 1$ and a local maximum for $x = 3$

Explanation:

We have:

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

the function is defined in all of $\mathbb{R}$ as ${x}^{2} + 3 > 0 \forall x$

We can identify the critical points by finding where the first derivative equals zero:

$f ' \left(x\right) = \frac{4 x}{{x}^{2} + 3} - 1 = - \frac{{x}^{2} - 4 x + 3}{{x}^{2} + 3}$

$- \frac{{x}^{2} - 4 x + 3}{{x}^{2} + 3} = 0$

${x}^{2} - 4 x + 3 = 0$

$x = 2 \pm \sqrt{4 - 3} = 2 \pm 1$

so the critical points are:

${x}_{1} = 1$ and ${x}_{2} = 3$

Since the denominator is always positive, the sign of $f ' \left(x\right)$ is the opposite of the sign of the numerator $\left({x}^{2} - 4 x + 3\right)$

Now we know that a second order polynomial with positive leading coefficient is positive outside the interval comprised between the roots and negative in the interval between the roots, so that:

$f ' \left(x\right) < 0$ for $x \in \left(- \infty , 1\right)$ and $x \in \left(3 , + \infty\right)$
$f ' \left(x\right) > 0$ for $x \in \left(1 , 3\right)$

We have then that $f \left(x\right)$ is decreasing in $\left(- \infty , 1\right)$, increasing in $\left(1 , 3\right)$, and again decreasing in $\left(3 , + \infty\right)$, so that ${x}_{1} = 1$ must be a local minimum and ${x}_{2} = 3$ must be a local maximum.

graph{2ln(x^2+3) -x [-1.42, 8.58, -0.08, 4.92]}