Question

which is the graph of $f(x) = \sqrt{x}$?

Explanation:

Step1: Analyze the function \( f(x)=\sqrt{x} \)

The domain of \( f(x)=\sqrt{x} \) is \( x\geq0 \) (since we can't take the square root of a negative number in real numbers), and the range is \( y\geq0 \) (because the square root of a non - negative number is non - negative). Also, let's find some key points: when \( x = 0 \), \( f(0)=\sqrt{0}=0 \); when \( x = 1 \), \( f(1)=\sqrt{1}=1 \); when \( x = 4 \), \( f(4)=\sqrt{4}=2 \).

Step2: Analyze each graph

• First graph: This is a parabola opening upwards with vertex at the origin, but it is symmetric about the y - axis (\( y = x^{2} \) graph), not the graph of \( y=\sqrt{x} \).
• Second graph: The domain starts at \( x = 0 \), when \( x = 0 \), \( y = 0 \); when \( x = 1 \), \( y = 1 \); when \( x = 4 \), \( y = 2 \). The shape is a curve that starts at the origin and increases slowly, which matches the behavior of \( y=\sqrt{x} \) (since the derivative \( f^\prime(x)=\frac{1}{2\sqrt{x}} \) is positive and decreasing as \( x \) increases, so the function is increasing at a decreasing rate).
• Third graph: This graph is steeper, and when \( x = 4 \), if we assume the graph is \( y = x^{2} \) for \( x\geq0 \), \( y = 16 \) at \( x = 4 \), which is not the case for \( y=\sqrt{x} \).
• Fourth graph: This graph has points with negative \( y \) - values for positive \( x \), but \( y=\sqrt{x}\geq0 \) for \( x\geq0 \), so this graph is not of \( y = \sqrt{x} \).

Answer:

The second graph (the one in the middle - left among the four graphs, with the curve starting at (0,0), passing through (1,1) and (4,2) and increasing at a decreasing rate)