Question

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Explanation:

Step1: Analyze the domain of the graph

The graph is defined for \( x > 0 \) (since it starts near \( x = 0 \) on the right). Let's check the domain of each function:

• \( f(x)=3 + \log_2 x \): Domain \( x>0 \) (logarithm of \( x \), \( x>0 \))
• \( f(x)=\log_2(x - 4) \): Domain \( x - 4>0\Rightarrow x>4 \) (does not match the graph's domain as graph is near \( x = 0 \))
• \( f(x)=-5+\log_2 x \): Domain \( x>0 \)
• \( f(x)=-\log_2(x + 5) \): Domain \( x+5>0\Rightarrow x>-5 \) (but the graph is for \( x>0 \), and the shape: let's check the y - intercept or a point)

Step2: Check the y - value at \( x = 1 \)

For \( f(x)=3+\log_2 x \), when \( x = 1 \), \( \log_2 1=0 \), so \( f(1)=3 + 0=3 \). The graph at \( x = 1 \) (near \( x = 1 \)) seems to have a y - value around 3? Wait, let's check the graph: when \( x = 1 \), the curve is at \( y = 3 \)? Wait, the graph passes near \( x = 1 \), \( y = 3 \)? Wait, let's check the other functions:

• For \( f(x)=-5+\log_2 x \), when \( x = 1 \), \( f(1)=-5+0=-5 \) (not matching the graph's y - value at \( x = 1 \))
• For \( f(x)=-\log_2(x + 5) \), when \( x = 1 \), \( f(1)=-\log_2(6)\approx - 2.58 \) (not matching)
• For \( f(x)=\log_2(x - 4) \), when \( x = 5 \), \( f(5)=\log_2(1)=0 \), but the graph at \( x = 5 \) is at \( y\approx5 \), not 0.
• For \( f(x)=3+\log_2 x \), when \( x = 1 \), \( y = 3 \); when \( x = 2 \), \( \log_2 2 = 1\), so \( y=3 + 1=4 \); when \( x = 4 \), \( \log_2 4=2\), so \( y=3 + 2=5 \). This matches the increasing trend of the graph (as \( x \) increases, \( y \) increases) and the domain \( x>0 \).

Step3: Eliminate other options

• \( f(x)=\log_2(x - 4) \): Domain \( x>4 \), graph starts at \( x\approx0 \), so eliminate.
• \( f(x)=-5+\log_2 x \): At \( x = 1 \), \( y=-5 \), graph at \( x = 1 \) is \( y\approx3 \), eliminate.
• \( f(x)=-\log_2(x + 5) \): This function is decreasing (since there's a negative sign in front of log), but the graph is increasing, so eliminate.

Answer:

\( f(x)=3+\log_2 x \) (the first option: \( \boldsymbol{f(x)=3 + \log_2 x} \))