The answer I got from my book is positive. However, I am wondering what if we defined:

$g(x)=x^2$ and $f(x)=\sqrt{x}$

Then $h(x)=f(g(x))=x$, but $y=x$ is and odd function!

so I am very confused now, could anybody help me?

  • 3
    $\begingroup$ Incorrect, $h(-1) = \sqrt{(-1)^2} = \sqrt{1} = 1$. $\endgroup$ Aug 12, 2022 at 5:14
  • 7
    $\begingroup$ $\sqrt{x^2}$ of course isn't $x$ but $\left | x \right |$ which is indeed an even function. $\endgroup$ Aug 12, 2022 at 5:15

1 Answer 1


(A) We are given that $g(-x)=g(x)$ (Even)

Let $h(x)=f(g(x))$ , then $h(-x)=f(g(-x))=f(g(x))=h(x)$ (Even)

(B) Where you are going wrong:

$\sqrt{x^{2}}=|x|$ where we have to take the Positive root.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .