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

1 Answer
Mar 4, 2018

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.