Locate all inflection points of $f ( x ) = \frac { x } { x ^ { 2 } + 1 }$ .
A) $\left( - \sqrt { 3 } , - \frac { \sqrt { 3 } } { 4 } \right) , ( 0,0 ) , \left( \sqrt { 3 } , \frac { \sqrt { 3 } } { 4 } \right)$
B) $( 0,0 ) , \left( 1 , \frac { 1 } { 2 } \right) , \left( - 1 , - \frac { 1 } { 2 } \right)$
C) (0, 0)
D) $( 0,0 ) , \left( - 1 , - \frac { 1 } { 2 } \right)$

Question

Quizplus · Accepted Answer

To find the inflection points of $f(x) = \frac{x}{x^2 + 1}$, we need to find where the second derivative of $f(x)$ changes sign, indicating a change in concavity. The first derivative $f'(x)$ is $\frac{1 - x^2}{(x^2 + 1)^2}$, and the second derivative $f''(x)$ is $\frac{-2x(x^2 - 3)}{(x^2 + 1)^3}$. Setting $f''(x) = 0$ gives $x = 0$ and $x = \pm \sqrt{3}$, which are the points where the concavity might change. Evaluating the sign of $f''(x)$ around these points confirms that the concavity changes at $x = 0$ and $x = \pm \sqrt{3}$, making them inflection points. The corresponding $y$-values are found by substituting these $x$-values back into $f(x)$, yielding the points $\left( - \sqrt { 3 } , - \frac { \sqrt { 3 } } { 4 } \right)$, $(0,0)$, and $\left( \sqrt { 3 } , \frac { \sqrt { 3 } } { 4 } \right)$.

Locate All Inflection Points Of f(x)=xx2+1f ( x ) = \frac { x } { x ^ { 2 } + 1 }f(x)=x2+1x​

Locate All Inflection Points Of $f ( x ) = \frac { x } { x ^ { 2 } + 1 }$