1. The problem states that $f'(3)$ is undefined and $f''(x) > 0$ for all $x \neq 3$. This means the function has a critical point (likely a cusp or sharp corner) at $x=3$, and is concave up everywhere else on the interval $[1,5]$. 2. Since $f''(x) > 0$ for $x \neq 3$, the graph must be concave up on both sides of $x=3$, excluding the point $x=3$ itself where the first derivative is undefined. 3. A cusp or sharp corner at $x=3$ implies the graph is continuous but the slope does not exist at that point (first derivative undefined). 4. Analyzing the graph descriptions: - (a) Concave up with minimum between 3 and 4 fits $f''(x) > 0$ but doesn't address $f'(3)$ undefined clearly. - (b) "V" shape with apex at $x=3$ indicates a sharp corner, $f'(3)$ undefined, and concave up sides. - (c) Gradually increasing and concave up would imply $f'(3)$ exists, contradicting the undefined derivative. - (d) Peak at $x=3$ and concave down contradicts $f''(x) > 0$. 5. Therefore, graph (b) best represents the function continuous on [1,5] with $f'(3)$ undefined and $f''(x) > 0$ when $x \neq 3$. Final answer: **(b)**