Question

identify the graph of $g(x) = x^3 + x + 3$.

Explanation:

Step1: Analyze the function's degree and leading coefficient

The function \( g(x) = x^3 + x + 3 \) is a cubic function (degree 3) with a leading coefficient of \( 1 \) (positive). For cubic functions \( ax^3+bx^2 + cx + d \), when \( a>0 \), as \( x\to\infty \), \( g(x)\to\infty \) and as \( x\to-\infty \), \( g(x)\to-\infty \). So we can eliminate graphs where the end - behaviors are reversed (e.g., as \( x\to\infty \), \( y\to-\infty \) and as \( x\to-\infty \), \( y\to\infty \)). This eliminates the graphs with "right - down, left - up" end - behaviors.

Step2: Find the derivative to check for critical points

The derivative of \( g(x) \) is \( g^\prime(x)=3x^{2}+1 \). Since \( x^{2}\geq0 \) for all real \( x \), \( 3x^{2}+1\geq1>0 \) for all real \( x \). This means the function \( g(x) \) is always increasing (no local maxima or minima). So we can eliminate graphs that have local maxima or minima (the graphs with "hills" or "valleys" in between).

Step3: Check the y - intercept

The y - intercept of a function \( y = g(x) \) is found by setting \( x = 0 \). For \( g(x)=x^{3}+x + 3 \), when \( x = 0 \), \( g(0)=0^{3}+0 + 3=3 \). So the graph should pass through the point \( (0,3) \).

Looking at the remaining graphs (after eliminating those with wrong end - behaviors and those with local extrema), the top - left graph has the correct end - behaviors (up on the right, down on the left), no local extrema (since the function is always increasing), and passes through \( (0,3) \).

Answer:

The top - left graph (the first graph in the top - left corner of the given set of graphs)