What are the points of inflection of #f(x)=4 / (2x^2#?

Question

1117 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-08T04:41:35

No points of inflection.

We have: #f(x) = (4) / (2 x^(2))#

First, let's evaluate the second derivative of this function using the "quotient rule":

#=> f'(x) = ((2 x^(2) cdot 0) - (4 cdot 4 x)) / ((2 x^(2))^(2))#

#=> f'(x) = - (16 x) / (4 x^(4))#

#=> f'(x) = - (4) / (x^(3))#

#=> f''(x) = ((x^(3) cdot 0) - (- 4 cdot 3 x^(2))) / ((x^(3))^(2))#

#=> f''(x) = (12 x^(2)) / (x^(6))#

#=> f''(x) = (12) / (x^(4))#

Then, to determine the points of inflection, we need to set the second derivative equal to zero, and then solve for #x#:

#=> f''(x) = 0#

#=> (12) / (x^(4)) = 0#

#=> 12 ne 0#

#therefore# no real solutions

Therefore, there are no points of inflection for #f(x)#.