Exercise 1.1.81 (81-86)

Question

Determine whether $f$ is even, odd, or neither. You may wish to use a graphing calculator or computer to check your answer visually.

Khagan Gulusoy · Accepted Answer

To find out whether a function $f$ is odd or even, we must first find $f(-x)$ %for all $x$ that belong to the domain of the function
. If $f(-x)=f(x),$ the function $f$ is even. Else if $f(-x)=-f(x),$ the function is odd, otherwise the function is neither even nor odd.
\begin{enumerate}
    \item [81.] $$f(-x)=\frac{-x}{(-x)^2+2}=\frac{-x}{x^2+2}=-\frac{x}{x^2+2}=-f(x).$$
    Thus, $f$ is odd.
    \begin{figure}[H] \centering
    \begin{tikzpicture}
    \begin{axis}[axis lines=middle, unit vector ratio=1 1 1, xmin=-3.2, xmax=3.2, ymin=-1, ymax=1,ticks=none, xlabel={$x$}, ylabel={$y$}, x label style = {at={(ticklabel cs:0.9)},anchor=south,above=4.8mm}, y label style={anchor=north, above=3mm}]
         \addplot[domain=-3:3, samples=50, mark=none, color=red, ultra thick] {x/(x^2+1)};          
    \end{axis}    
    \end{tikzpicture}
    \end{figure}
    
    \item[82.] $$f(-x)=\frac{(-x)^2}{(-x)^4+1}=\frac{x^2}{x^4+1}=f(x).$$
    \begin{figure}[H] \centering
    \begin{tikzpicture}
    \begin{axis}[axis lines=middle, unit vector ratio=1 1 1, xmin=-3.2, xmax=3.2, ymin=-0.2, ymax=1,ticks=none, xlabel={$x$}, ylabel={$y$}, x label style = {at={(ticklabel cs:0.9)},anchor=south,above=4.8mm}, y label style={anchor=north, above=3mm}]
         \addplot[domain=-3:3, samples=50, mark=none, color=red, ultra thick] {x^2/(x^4+1)};          
    \end{axis}    
    \end{tikzpicture}
    \end{figure}
    Since $f(-x)=f(x)$, $f$ is even.
    \item[83.] $$f(-x)=\frac{-x}{-x+1}=\frac{x}{x-1}\implies f(-x)
eq f(x) 	ext{ and } f(-x)
eq -f(x).$$
    Therefore, $f$ is neither odd nor even.
    \begin{figure}[H] \centering
    \begin{tikzpicture}
    \begin{axis}[axis lines=middle, xmin=-3.2, xmax=3.2, ticks=none, xlabel={$x$}, ylabel={$y$}, x label style = {at={(ticklabel cs:0.9)},anchor=south,above=4.8mm}, y label style={anchor=north, above=3mm}]
         \addplot[domain=-3:-1.1, samples=50, mark=none, color=red, ultra thick] {x/(x+1)};
         \addplot[domain=-0.9:3, samples=50, mark=none, color=red, ultra thick] {x/(x+1)};
    \end{axis}    
    \end{tikzpicture}
    \end{figure}
    \item[84.] $$f(-x)=-x|-x|=-x|x|=-f(x).$$
    Thus, $f$ is odd.
    \begin{figure}[H] \centering
    \begin{tikzpicture}
    \begin{axis}[axis lines=middle, unit vector ratio=1 1 1, xmin=-3.2, xmax=3.2, ymin=-4, ymax=4,ticks=none, xlabel={$x$}, ylabel={$y$}, x label style = {at={(ticklabel cs:0.9)},anchor=south,above=4.8mm}, y label style={anchor=north, above=3mm}]
         \addplot[domain=-3:3, samples=50, mark=none, color=red, ultra thick] {x*abs(x)};          
    \end{axis}    
    \end{tikzpicture}
    \end{figure}
    \item[85.] $$f(-x)=1+3(-x)^2-(-x)^4=1+3x^2-x^4=f(x).$$
    Therefore, function $f$ is an even function.
    \begin{figure}[H] \centering
    \begin{tikzpicture}
    \begin{axis}[axis lines=middle, unit vector ratio=1 1 1, xmin=-3.2, xmax=3.2, ymin=-4, ymax=4,ticks=none, xlabel={$x$}, ylabel={$y$}, x label style = {at={(ticklabel cs:0.9)},anchor=south,above=4.8mm}, y label style={anchor=north, above=3mm}]
         \addplot[domain=-3:3, samples=50, mark=none, color=red, ultra thick] {1+3*x^2-x^4};          
    \end{axis}    
    \end{tikzpicture}
    \end{figure}
    \item [86.] $$f(-x)=1+3(-x)^3-(-x)^5=1-3x^3+x^5\implies f(-x)
eq f(x) 	ext{ and } f(-x)
eq -f(x).$$
    Therefore, $f$ is neither odd nor even.
    \begin{figure}[H] \centering
    \begin{tikzpicture}
    \begin{axis}[axis lines=middle, unit vector ratio=1 1 1, xmin=-3.2, xmax=3.2, ymin=-4, ymax=4,ticks=none, xlabel={$x$}, ylabel={$y$}, x label style = {at={(ticklabel cs:0.9)},anchor=south,above=4.8mm}, y label style={anchor=north, above=3mm}]
         \addplot[domain=-3:3, samples=50, mark=none, color=red, ultra thick] {1+3*x^3-x^5};          
    \end{axis}    
    \end{tikzpicture}
    \end{figure}
\end{enumerate}

Exercise 1.1.81 (81-86)

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Exercise 1.1.81 (81-86)

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