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avatar+159 

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.
 

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

 May 1, 2021

Best Answer 

 #1
avatar+9299 
+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 laugh

 May 1, 2021
 #1
avatar+9299 
+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 laugh

hectictar May 1, 2021
 #2
avatar+159 
+3

ok thanks, i'll check

VooFIX  May 1, 2021
 #3
avatar+159 
+3

Yea, you were correct, Cubed root 3 was odd, not neither. The answer was 1,3,4

VooFIX  May 1, 2021

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