Question

if $f(x) = x^2 + 3$, $g(x) = x - 2$, and $h(x) = \sqrt{x}$, then
$(f \circ g)(x) = $
$(g \circ f)(x) = $
$(h \circ g)(x) = $
give answers in simplified form
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Explanation:

Step1: Substitute $g(x)$ into $f(x)$

$(f \circ g)(x) = f(g(x)) = (x-2)^2 + 3$

Step2: Expand and simplify the expression

$(x-2)^2 + 3 = x^2 -4x +4 +3 = x^2 -4x +7$

Step3: Substitute $f(x)$ into $g(x)$

$(g \circ f)(x) = g(f(x)) = (x^2 + 3) - 2$

Step4: Simplify the expression

$(x^2 + 3) - 2 = x^2 +1$

Step5: Substitute $g(x)$ into $h(x)$

$(h \circ g)(x) = h(g(x)) = \sqrt{x-2}$

Answer:

$(f \circ g)(x) = x^2 -4x +7$
$(g \circ f)(x) = x^2 +1$
$(h \circ g)(x) = \sqrt{x-2}$