Question

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

Answer 1

The inflection points for this function are `+-[2^(1/2)]`

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''(x)=0`. A sufficient condition requires `f^('')(x+epsilon)` and `f^('')(x-epsilon)` 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''(x)=0`.

`f(x)= 3ln(x^2+2) -2x`
`rArr f'(x)= [3(2x)/(x^2+2)]-2` (Using chain Rule)

`rArr f''(x)=[6(x^2+2) -12x^2]/(x^2+2)=[-6x^2 + 12]/(x^2+2)^2` ----(i)

Putting f''(x)=0, we get:
`-6x^2+12=0` `rArr x=+-2^(1/2)`

Now, to check if `x=+-2^(1/2)` really are the inflection points of this function, we check neighboring points of `+-2^(1/2)`.
Substitute the following values in expression (i) and we get;
`f''(1) = 2/3=positive`
`f''(2) = -1/3=` negative
Hence, `f''(x)` has opposite signs at points lying very close to and to the left and right of `2^(1/2)`. So `2^(1/2)` is an inflection point.

Similarly, check;
`f''(-1) = 2/3=positive`
`f''(-2) = -1/3=` negative
By the same argument, `-2^(1/2)` is also an inflection point.