Question

Let $f\left(x\right)=|x-2|+|x-3|+|x-4|$ and $g\left(x\right)=f(x+1)$. Then

• A. $g\left(x\right)$ is an even function
• B. $g\left(x\right)$ is an odd function
• C. $g\left(x\right)$ is neither even nor odd
• D. $g\left(x\right)$ is periodic

Answer 1

Answer: (C)

$g\left(x\right)$ is neither even nor odd

$f\left(x\right)=|x-2|+|x-3|+|x-4|$

$f\left(\right(x+1\left)\right)=|x+1-2|+|x+1-3|+|x+1-4|$

$g\left(x\right)=|x-1|+|x-2|+|x-3|$

$g(-x)=|-x-1|+|-x-2|+|-x-3|=|x+1|+|x+2|+|x+3|\ne g\left(x\right),also\ne -g\left(x\right)$

Hence it will neither be an odd or an even function.