Among the following functions, let **a** be the number of functions that are even, let **b** be the number of functions that are odd, and let **c** be the number of functions that are neither even nor odd. Enter your answer in the form "**a,b,c** ".

I've tried to solve it and got 1,2,5 but it's incorrect. Where the log one is Even, the 1st equation and the 6x^5 is odd, and the rest are neither.

\(\begin{align*} y &= \frac{x}{x^2 - 1} \\ y &= \frac{x}{x - 1} \\ y &= 2x - 11 \\ y &= \log_3 (x^2 - 1) \\ y &= 2^{-x} \\ y &= 6x^5 + x + \frac{2}{x^3} \\ y &= \sqrt[3]{x} \\ y &= \lfloor x \rfloor \end{align*}\)

VooFIX May 1, 2021

#1**+2 **

I think \(y=\sqrt[3]{x}\) is not neither.

\(f(x)=\sqrt[3]{x}\\~\\ f(-x)\ =\ \sqrt[3]{-x}\ =\ \sqrt[3]{-1\cdot x}\ =\ \sqrt[3]{-1}\cdot\sqrt[3]{x}\ =\ -1\cdot\sqrt[3]{x}\ =\ -\sqrt[3]{x}\ =\ -f(x)\)

(It is a little confusing because when I check that in WolframAlpha, it only agrees when it uses the "real-valued root")

Otherwise I agree with your answers

hectictar May 1, 2021

#1**+2 **

Best Answer

I think \(y=\sqrt[3]{x}\) is not neither.

\(f(x)=\sqrt[3]{x}\\~\\ f(-x)\ =\ \sqrt[3]{-x}\ =\ \sqrt[3]{-1\cdot x}\ =\ \sqrt[3]{-1}\cdot\sqrt[3]{x}\ =\ -1\cdot\sqrt[3]{x}\ =\ -\sqrt[3]{x}\ =\ -f(x)\)

(It is a little confusing because when I check that in WolframAlpha, it only agrees when it uses the "real-valued root")

Otherwise I agree with your answers

hectictar May 1, 2021