Let f(x)=sqrt{x^{2}-3x+2} and g(x)=sqrt{x} be | vedclass

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(A) For $f \circ g(x) = f(g(x)) = \sqrt{(\sqrt{x})^2 - 3\sqrt{x} + 2} = \sqrt{x - 3\sqrt{x} + 2}$.For the domain $S$, we require $x \ge 0$ and $x - 3\

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$S=T$

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(A) For $f \circ g(x) = f(g(x)) = \sqrt{(\sqrt{x})^2 - 3\sqrt{x} + 2} = \sqrt{x - 3\sqrt{x} + 2}$.For the domain $S$, we require $x \ge 0$ and $x - 3\sqrt{x} + 2 \ge 0$.Let $u = \sqrt{x}$, then $u^2 - 3u + 2 \ge 0 \implies (u-1)(u-2) \ge 0$.This gives $u \le 1$ or $u \ge 2$, so $\sqrt{x} \le 1$ or $\sqrt{x} \ge 2$.Thus, $x \in [0, 1] \cup [4, \infty)$, so $S = [0, 1] \cup [4, \infty)$.For $g \circ f(x) = g(f(x)) = \sqrt{\sqrt{x^2 - 3x + 2}}$.For the domain $T$, we require $x^2 - 3x + 2 \ge 0 \implies (x-1)(x-2) \ge 0$.This gives $x \le 1$ or $x \ge 2$, so $T = (-\infty, 1] \cup [2, \infty)$.Now, $S \cap T = ([0, 1] \cup [4, \infty)) \cap ((-\infty, 1] \cup [2, \infty)) = [0, 1] \cup [4, \infty)$.Since $S \cap T = S$, and $S$ is a union of two disjoint intervals, the intersection is not a single interval.

Let $f(x)=\sqrt{x^{2}-3x+2}$ and $g(x)=\sqrt{x}$ be two given functions. If $S$ is the domain of $f \circ g$ and $T$ is the domain of $g \circ f$, then:

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