# What are the points of inflection of f(x)=3ln(x^(2) +2) -2x ?

Mar 4, 2018

The inflection points for this function are $\pm \left[{2}^{\frac{1}{2}}\right]$

#### Explanation:

To find out inflection points for any function, the second derivative test is also useful. A necessary condition for x to be an inflection point is $f ' ' \left(x\right) = 0$. A sufficient condition requires ${f}^{' '} \left(x + \epsilon\right)$ and ${f}^{' '} \left(x - \epsilon\right)$ to have opposite signs in the neighborhood of x (Bronshtein and Semendyayev 2004, p. 231).

Using the above test, we go ahead solving this question by differentiating the function twice, and then finding the values of $x$ for which $f ' ' \left(x\right) = 0$.

$f \left(x\right) = 3 \ln \left({x}^{2} + 2\right) - 2 x$
$\Rightarrow f ' \left(x\right) = \left[3 \frac{2 x}{{x}^{2} + 2}\right] - 2$ (Using chain Rule)

$\Rightarrow f ' ' \left(x\right) = \frac{6 \left({x}^{2} + 2\right) - 12 {x}^{2}}{{x}^{2} + 2} = \frac{- 6 {x}^{2} + 12}{{x}^{2} + 2} ^ 2$ ----(i)

Putting f''(x)=0, we get:
$- 6 {x}^{2} + 12 = 0$ $\Rightarrow x = \pm {2}^{\frac{1}{2}}$

Now, to check if $x = \pm {2}^{\frac{1}{2}}$ really are the inflection points of this function, we check neighboring points of $\pm {2}^{\frac{1}{2}}$.
Substitute the following values in expression (i) and we get;
$f ' ' \left(1\right) = \frac{2}{3} = p o s i t i v e$
$f ' ' \left(2\right) = - \frac{1}{3} =$ negative
Hence, $f ' ' \left(x\right)$ has opposite signs at points lying very close to and to the left and right of ${2}^{\frac{1}{2}}$. So ${2}^{\frac{1}{2}}$ is an inflection point.

Similarly, check;
$f ' ' \left(- 1\right) = \frac{2}{3} = p o s i t i v e$
$f ' ' \left(- 2\right) = - \frac{1}{3} =$ negative
By the same argument, $- {2}^{\frac{1}{2}}$ is also an inflection point.