1. **Problem Statement:** We are given the graph of the first derivative $f'(x)$ of a continuous function $f$ on $\mathbb{R}$. We need to identify which statement among (a), (b), (c), and (d) is incorrect. 2. **Recall Key Concepts:** - A **critical point** of $f$ occurs where $f'(x) = 0$ or $f'(x)$ is undefined. - An **inflection point** occurs where the concavity of $f$ changes, i.e., where $f''(x)$ changes sign. - The function $f$ is **convex up** (concave up) on intervals where $f''(x) > 0$. 3. **Analyze the Graph of $f'(x)$:** - The graph of $f'(x)$ is positive and steeply increasing near $x=0$. - There is a vertical asymptote near $x=0$ for $f'(x)$, indicating $f'(x)$ is not defined at $x=0$. - $f'(x)$ is positive on both sides of $0$ but not zero at $0$. 4. **Evaluate Each Statement:** (a) **Inflection point at $x=0$:** - Since $f'(x)$ is increasing steeply and has a vertical asymptote at $0$, $f''(x)$ changes sign around $0$. - This implies an inflection point at $x=0$. - So, (a) is **true**. (b) **$f$ is convex up on $(-\infty,0)$:** - Convex up means $f''(x) > 0$. - Since $f'(x)$ is increasing on $(-\infty,0)$, $f''(x) > 0$ there. - So, (b) is **true**. (c) **$f$ has a critical point at $x=0$:** - Critical points require $f'(0) = 0$ or $f'(0)$ undefined. - Here, $f'(x)$ has a vertical asymptote at $0$, so $f'(0)$ is undefined. - Thus, $x=0$ is a critical point. - So, (c) is **true**. (d) **$f$ is convex up on $(0, \infty)$:** - $f'(x)$ is positive but flattening out on $(0, \infty)$, meaning $f'(x)$ is decreasing or constant. - If $f'(x)$ is decreasing, then $f''(x) < 0$ on $(0, \infty)$. - So, $f$ is **not** convex up on $(0, \infty)$. - Therefore, (d) is **false**. 5. **Final Answer:** The wrong statement is (d). $$\boxed{\text{The wrong statement is (d).}}$$