1. The problem states that $f'(3)$ is undefined and $f''(x) > 0$ for all $x \neq 3$ on the interval $[1,5]$. 2. Since $f''(x) > 0$ for $x \neq 3$, the function $f$ is concave upward everywhere except possibly at $x=3$. 3. The derivative $f'(3)$ being undefined suggests a sharp corner or cusp at $x=3$. 4. A concave upward function with a sharp corner at $x=3$ resembles a "V" shape with the lowest point at $x=3$. 5. Among the given options, graph (b) shows a "V" shape concave upward curve starting at $(3, \text{lowest point})$. 6. Therefore, the curve representing the continuous function $f$ on $[1,5]$ is graph (b).