Use graphical methods to find any turning points of the graph of the function.
-$$f ( x ) = \frac { 1 } { x ^ { 2 } + 1 }$$
A) $( - 1,0.5 )$
B) $( - 1,0.5 ) , ( 0,1 ) , ( 1,0.5 )$
C) $( 1,0.5 )$
D) $( 0,1 )$

Question

Quizplus · Accepted Answer

The function $f(x) = \frac{1}{x^2 + 1}$ is symmetric about the y-axis because it is an even function (replacing $x$ with $-x$ does not change the function). The only turning point is at $x = 0$, where the function reaches its maximum value of $1$. This is because the denominator $x^2 + 1$ is minimized when $x = 0$, making the fraction itself maximized. There are no other turning points because the function is decreasing on both sides of $x = 0$ and does not have any other local maxima or minima.

Use Graphical Methods to Find Any Turning Points of the Graph