consider the functions.\n$f(x)=\\sqrt{x}$\n$g(x)=\\sqrt{x - 3}+1$\n$h(x)=\\sqrt{x + 1}-2$\nwhich statement…

Explanation:

Step1: Find minimum of (f(x))

The domain of (f(x)=\sqrt{x}) is (x\geq0). Since (\sqrt{x}\geq0), the minimum of (f(x)) occurs when (x = 0), and (f(0)=0). The minimum - point is ((0,0)).

Step2: Find minimum of (g(x))

The domain of (g(x)=\sqrt{x - 3}+1) is (x\geq3). The square - root function (\sqrt{x - 3}\geq0). The minimum of (g(x)) occurs when (\sqrt{x - 3}=0), i.e., when (x = 3). Then (g(3)=\sqrt{3 - 3}+1=1). The minimum - point is ((3,1)).

Step3: Find minimum of (h(x))

The domain of (h(x)=\sqrt{x + 1}-2) is (x\geq - 1). The square - root function (\sqrt{x + 1}\geq0). The minimum of (h(x)) occurs when (\sqrt{x + 1}=0), i.e., when (x=-1). Then (h(-1)=\sqrt{-1 + 1}-2=-2). The minimum - point is ((-1,-2)).

Step4: Analyze the relative positions

The minimum of (f(x)) is at ((0,0)), the minimum of (g(x)) is at ((3,1)) (in the first quadrant), and the minimum of (h(x)) is at ((-1,-2)) (in the third quadrant). The minimum of (h(x)) is farther left and down from the minimums of (f(x)) and (g(x)).

Answer:

The minimum of (h(x)) is farther left and down from the minimums of (f(x)) and (g(x)).