Given $f(x)= 1- x^2$ and $g(x) = \sqrt x$
What is the domain of $f$ and $g$? My answer is that the domain of $f$ is all real $x$, and the domain of $g$ is all $x \ge 0$.
However, I am not sure if this is correct. The question does not specify that we are in the real vector space. Does this mean the domain includes even the complex numbers.
Also, my formula for the composition $f \circ g$ is $1-x$. I got this by putting $\sqrt x$ in $1- x^2$. Domain of this would be $x \ge 0$.
and $g \circ f$ is $\sqrt {1-x^2}$. Domain of this is $x \ge 1$.
Have I got this right or are they backwards?
If $f$ and $g$ are real-valued functions of a real variable, then all your answers are correct except for the domain of $g \circ f$. Observe that if $x > 1$, then $1 - x^2 < 0$, so $\sqrt{1 - x^2}$ is not real-valued. Since $g \circ f = \sqrt{1 - x^2}$ and the domain of $f$ is the set of all real numbers, the only restriction is that $1 - x^2 \geq 0 \iff 1 \geq x^2 \iff 1 \geq |x|$. Thus, the domain of $g \circ f$ is $D_{g \circ f} = [-1, 1]$.