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

Question

Cameron Alexander · Accepted Answer

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)#.

Charles Arocha · Answer

undefined To find the points of inflection of $ f(x) = \frac{4}{2x^2} $, we need to find where the concavity changes.

First, find the second derivative of $ f(x) $:
$$ f''(x) = \frac{d^2}{dx^2} \left( \frac{4}{2x^2} \right) $$

$$ f''(x) = \frac{d}{dx} \left( -\frac{8}{x^3} \right) $$

$$ f''(x) = \frac{24}{x^4} $$

There are no points where $ f''(x) $ is equal to zero since the denominator can't be zero. Therefore, there are no points of inflection for this function.