Question

Is the function $f\left(x\right)=\left[\right(x+1{)}^2{]}^{1/3}+\left[\right(x-1{)}^2{]}^{1/3}$ odd, even or neither?

• A. Even
• B. Odd
• C. Neither even nor odd
• D. Even for some values of $x$

Answer 1

The correct option is A Even

Given, $f\left(x\right)=\left[\right(x+1{)}^2{]}^{1/3}+\left[\right(x-1{)}^2{]}^{1/3}$
Now $f(-x)=\left[\right(-x+1{)}^2{]}^{1/3}+\left[\right(-x-1{)}^2{]}^{1/3}=\left[\right(x-1{)}^2{]}^{1/3}+\left[\right(x+1{)}^2{]}^{1/3}=\left[\right(x+1{)}^2{]}^{1/3}+\left[\right(x-1{)}^2{]}^{1/3}$
Clearly $f\left(x\right)=f(-x)$, Hence $f\left(x\right)$ is an even function.