1. The problem states that $f$ is continuous with $f(0) = 3$, $f'(2) = f'(-2) = 0$, and $f'(x) > 0$ for $-2 < x < 2$. 2. The condition $f'(2) = f'(-2) = 0$ means the slope of the tangent line to the curve is zero at $x = -2$ and $x = 2$. 3. The condition $f'(x) > 0$ for $-2 < x < 2$ means the function is strictly increasing between $x = -2$ and $x = 2$. 4. Since $f(0) = 3$, the curve passes through the point $(0,3)$. 5. Among the given graphs: - Graph (a) has slope negative between $-2$ and $0$ and positive between $0$ and $2$, so $f'(x)$ is not positive on the entire interval $(-2,2)$. - Graph (b) has slope negative between $-2$ and $2$, so $f'(x)$ is not positive on $(-2,2)$. - Graph (c) has slope positive between $-2$ and $2$, matching $f'(x) > 0$ on $(-2,2)$. - Graph (d) has slope mostly negative or zero between $-2$ and $2$, so $f'(x)$ is not positive on $(-2,2)$. 6. Therefore, the graph that represents the function $f$ with the given conditions is graph (c).