Question

consider the functions.
f(x) = \sqrt{x}
g(x) = \sqrt{x - 3} + 1
h(x) = \sqrt{x + 1} - 2
which statement compares the relative locations of the minimums of the functions?
\bigcirc the minimums of g(x) and h(x) are both in the first quadrant.
\bigcirc the minimums of g(x) and h(x) are both in the third quadrant.
\bigcirc the minimum of h(x) is farther right and up from the minimums of f(x) and g(x).
\bigcirc the minimum of h(x) is farther left and down from the minimums of f(x) and g(x).

Explanation:

Step1: Find minimum of \( f(x) \)

The function \( f(x)=\sqrt{x} \) has domain \( x\geq0 \). The minimum occurs at \( x = 0 \), so \( f(0)=\sqrt{0}=0 \). The minimum point is \( (0,0) \).

Step2: Find minimum of \( g(x) \)

For \( g(x)=\sqrt{x - 3}+1 \), the domain is \( x\geq3 \). At \( x = 3 \), \( g(3)=\sqrt{3 - 3}+1=0 + 1=1 \). The minimum point is \( (3,1) \).

Step3: Find minimum of \( h(x) \)

For \( h(x)=\sqrt{x + 1}-2 \), the domain is \( x\geq - 1 \). At \( x=-1 \), \( h(-1)=\sqrt{-1 + 1}-2=0-2=-2 \). The minimum point is \( (-1,-2) \).

Step4: Compare locations

• The minimum of \( f(x) \) is \( (0,0) \), \( g(x) \) is \( (3,1) \), and \( h(x) \) is \( (-1,-2) \).
• Comparing \( h(x) \)'s minimum \( (-1,-2) \) with \( f(x) \)'s \( (0,0) \) and \( g(x) \)'s \( (3,1) \): \( -1<0 \) and \( -1<3 \) (farther left), \( -2<0 \) and \( -2<1 \) (farther down).

Answer:

The minimum of \( h(x) \) is farther left and down from the minimums of \( f(x) \) and \( g(x) \). (The last option among the given choices)