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**Prove It** Ah, no we can't, at least not if we're assuming $\displaystyle \displaystyle f: \mathbf{R} \to \mathbf{R}, f(x) = \sqrt{x}$ and $\displaystyle \displaystyle g: \mathbf{R} \to \mathbf{R}, g(x) = \sqrt{x}$.

$\displaystyle \displaystyle \sqrt{x^3}$ is not defined for any $\displaystyle \displaystyle x < 0$, so that means $\displaystyle \displaystyle f \circ g(x)$ does not exist for any $\displaystyle \displaystyle g(x)$ which are negative. In this case for $\displaystyle f \circ g(x)$ to exist, we require restricting the domain of $\displaystyle \displaystyle g(x)$ so that the range will be $\displaystyle \displaystyle [0, \infty)$.