Is the following function even, odd, or neither? f(x)=(x)/(x^(2)+1) Choose 1 answer: (A) Even (B) Odd (c) Neither

Define function and find f(-x): Define the function f(x) and find f(-x). The function given is f(x)=(x)/(x^2+1). To determine if the function is even, odd, or neither, we need to evaluate f(-x) and compare it to f(x). Calculate f(-x) by substituting -x for x in the function. f(-x)=(-x)/((-x)^2+1)Calculate f(-x): Simplify f(-x). Simplify the expression by noting that (-x)^2 is the same as x^2 because the square of a negative number is positive. f(-x)=(-x)/(x^2+1)Simplify f(-x): Compare f(x) and f(-x). We have f(x)=(x)/(x^2+1) and f(-x)=(-x)/(x^2+1). Notice that f(-x) is not equal to f(x) because the numerator changes sign, and f(-x) is not equal to -f(x) because only the numerator changes sign while the denominator remains the same. Therefore, the function is neither even nor odd.

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Is the following function even, odd, or neither?

f(x)=(x)/(x^(2)+1)
Choose 1 answer:
(A) Even
(B) Odd
(c) Neither

Is the following function even, odd, or neither? $\newline$ $f(x)=\frac{x}{x^{2}+1}$ $\newline$ Choose $1$ answer: $\newline$ (A) Even $\newline$ (B) Odd $\newline$ (C) Neither

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Algebra 2

Even and odd functions

Full solution

Q. Is the following function even, odd, or neither?

\newline

f(x)=\frac{x}{x^{2}+1}

\newline

Choose

1

answer:

\newline

(A) Even

\newline

(B) Odd

\newline

Define function and find f(-x): Define the function f(x) and find f(-x). $\newline$ The function given is $f(x)=\frac{x}{x^2+1}$ . To determine if the function is even, odd, or neither, we need to evaluate $f(-x)$ and compare it to $f(x)$ . $\newline$ Calculate $f(-x)$ by substituting $-x$ for $x$ in the function. $\newline$ $f(-x)=\frac{-x}{(-x)^2+1}$
Calculate $f(-x)$ : Simplify $f(-x)$ . $\newline$ Simplify the expression by noting that $(-x)^2$ is the same as $x^2$ because the square of a negative number is positive. $\newline$ $f(-x)=\frac{-x}{x^2+1}$
Simplify $f(-x)$ : Compare $f(x)$ and $f(-x)$ . $\newline$ We have $f(x)=\frac{x}{x^2+1}$ and $f(-x)=\frac{-x}{x^2+1}$ . $\newline$ Notice that $f(-x)$ is not equal to $f(x)$ because the numerator changes sign, and $f(-x)$ is not equal to $-f(x)$ because only the numerator changes sign while the denominator remains the same. $\newline$ Therefore, the function is neither even nor odd.