1. We are given that $f$ is continuous on $[a,b]$ and for every $x_1, x_2 \in [a,b]$, the inequality $f'(x_2) - f'(x_1) > 0$ holds when $x_2 > x_1$. This means the derivative $f'$ is strictly increasing on $[a,b]$. 2. Since $f'$ is increasing, its derivative $f''$ must be positive on the open interval $(a,b)$. That is, $$f''(x) > 0 \quad \text{for all} \quad x \in (a,b).$$ 3. The second derivative test tells us that if $f''(x) > 0$ on $(a,b)$, then the function $f$ is convex (concave upward) on $(a,b)$. In other words, the curve of $f$ is convex upward in the interval $]a,b[$. 4. Note that the condition on $f'$ increasing does not guarantee $f$ itself is increasing or decreasing, only its curvature (convexity). 5. Therefore, the correct conclusion is option (c): the curve of $f$ is convex upward on the interval $]a,b[$.