Let's check each function for **log-concavity**. Recall: A function \( f(x) \) is **log-concave** if \( \log f(x) \) is a concave function, i.e., \( \frac{d^2}{dx^2} (\log f(x)) \leq 0 \). --- ### 1. \( f(x) = \min\{2, x\} \) for \( x > 0 \) - For \( x < 2 \), \( f(x) = x \). For \( x \geq 2 \), \( f(x) = 2 \). - \( \log f(x) = \log x \) for \( x < 2 \) (concave), and constant for \( x \geq 2 \). - The minimum of two log-concave functions is **not always log-concave**. At \( x = 2 \), the function is not smooth, so **not log-concave everywhere**. --- ### 2. \( f(x) = e^{\max\{-3, -x^2\}} \) on \( (-\infty, \infty) \) - \( \log f(x) = \max\{-3, -x^2\} \). - For \( |x| \leq \sqrt{3} \), \( \log f(x) = -x^2 \) (concave). - For \( |x| \geq \sqrt{3} \), \( \log f(x) = -3 \) (constant, concave). - The maximum of concave functions is concave, so this function **is log-concave**. --- ### 3. \( f(x) = -x \log(x) \) on \( (0, 1) \) - For \( x \in (0,1) \), \( -x\log x > 0 \). - \( \log f(x) = \log(-x \log x) = \log x + \log(-\log x) \). - Take second derivative: - \( g(x) = \log x + \log(-\log x) \) - \( g'(x) = \frac{1}{x} + \frac{-1}{x \log x} \) - \( g''(x) = -\frac{1}{x^2} + \frac{1}{x^2 \log x} - \frac{1}{x^2 (\log x)^2} \) - For \( x \in (0,1) \), \( \log x < 0 \), so \( g''(x) < 0 \). - So, **log-concave**. --- ### 4. \( f(x) = \frac{x}{1+x} \) for \( x > 0 \) - \( \log f(x) = \log x - \log(1+x) \) - Take second derivative: - \( (\log x)' = 1/x \), \( (\log(1+x))' = 1/(1+x) \) - \( (\log x - \log(1+x))'' = -1/x^2 + 1/(1+x)^2 \) - For \( x > 0 \), \( 1/(1+x)^2 < 1/x^2 \), so the sum is **negative**. - So, **log-concave**. --- ### 5. \( f(x) = \log(1 + e^{-x}) \) on \( (-\infty, \infty) \) - \( \log f(x) = \log(\log(1 + e^{-x})) \) - Take derivative: - Let \( g(x) = \log(1 + e^{-x}) \) - \( g'(x) = \frac{-e^{-x}}{1 + e^{-x}} \) - \( g''(x) = \frac{e^{-x}}{1 + e^{-x}} - \frac{e^{-2x}}{(1 + e^{-x})^2} \) - But \( f(x) \) is known to be **log-convex**, not log-concave** for all \( x \). (Or, you can check the second derivative is positive for some \( x \).) --- ## **Summary** - 2. Log-concave - 3. Log-concave - 4. Log-concave **Final Answer:** \[ \boxed{3} \] functions are log-concave.

Let's analyze each function carefully to determine whether it is log-concave. Recall that a function \(f(x)\) is **log-concave** if \(\log f(x)\) is a concave function, i.e., its second derivative is less than or equal to zero: \(\frac{d^2}{dx^2} \log f(x) \leq 0\). --- ### 1. \(f(x) = \min\{2, x\}\) for \(x > 0\) **Step 1:** Understand \(f(x)\): - For \(x < 2\), \(f(x) = x\). - For \(x \geq 2\), \(f(x) = 2\). **Step 2:** Logarithm: - For \(x < 2\), \(\log f(x) = \log x\). - For \(x \geq 2\), \(\log f(x) = \log 2\) (constant). **Step 3:** Concavity of \(\log f(x)\): - For \(x < 2\), \(\log x\): - First derivative: \(\frac{1}{x}\). - Second derivative: \(-\frac{1}{x^2} \leq 0\) (concave). - For \(x \geq 2\), \(\log 2\), a constant (concave). - **However**, at \(x=2\), the function \(\log f(x)\) has a "kink" (not smooth), with a jump in the derivative, indicating **not smoothness**. Since the second derivative test applies to smooth functions, and the function isn't smooth at \(x=2\), the overall function **may not be log-concave everywhere**. **Conclusion:** Because of the non-smooth point at \(x=2\), \(f(x)\) is **not log-concave** on the entire domain. --- ### 2. \(f(x) = e^{\max\{-3, -x^2\}}\) on \((-\infty, \infty)\) **Step 1:** Express \(\log f(x)\): \[ \log f(x) = \max\{-3, -x^2\} \] **Step 2:** Behavior of \(\log f(x)\): - For \(|x| \leq \sqrt{3}\), \(-x^2 \geq -3\), so \(\log f(x) = -x^2\). - For \(|x| \geq \sqrt{3}\), \(-x^2 \leq -3\), so \(\log f(x) = -3\). **Step 3:** Is \(\log f(x)\) concave? - On \(|x| \leq \sqrt{3}\): - \(\log f(x) = -x^2\), which is **concave** (second derivative: \(-2 \leq 0\)). - On \(|x| \geq \sqrt{3}\): - \(\log f(x) = -3\), a constant (concave). **Step 4:** Check at the transition point \(x = \pm \sqrt{3}\): - The function is continuous and piecewise concave, and the maximum of two concave functions is **concave**. **Conclusion:** \(\log f(x)\) is concave on \((-\infty, \infty)\). **Therefore,** \(f(x)\) is **log-concave**. --- ### 3. \(f(x) = -x \log x\) for \(x \in (0,1)\) **Step 1:** Compute \(\log f(x)\): \[ \log f(x) = \log(-x \log x) = \log x + \log(-\log x) \] Note: Since \(x \in (0,1)\), \(\log x < 0\), and \(-\log x > 0\), so the expression is valid. **Step 2:** Derivatives: \[ g(x) = \log x + \log(-\log x) \] First derivative: \[ g'(x) = \frac{1}{x} + \frac{1}{-\log x} \cdot \left(-\frac{1}{x}\right) = \frac{1}{x} - \frac{1}{x \log x} = \frac{1}{x} \left(1 - \frac{1}{\log x}\right) \] Second derivative: \[ g''(x) = \frac{d}{dx} \left[\frac{1}{x} \left(1 - \frac{1}{\log x}\right)\right] \] Applying the product rule: \[ g''(x) = -\frac{1}{x^2} \left(1 - \frac{1}{\log x}\right) + \frac{1}{x} \cdot \frac{d}{dx} \left(- \frac{1}{\log x}\right) \] Compute \(\frac{d}{dx} \left(- \frac{1}{\log x}\right)\): \[ \frac{d}{dx} \left(- \frac{1}{\log x}\right) = - \frac{d}{dx} \left(\frac{1}{\log x}\right) = - \left(- \frac{1}{(\log x)^2} \cdot \frac{1}{x}\right) = \frac{1}{x (\log x)^2} \] Now, substitute back: \[ g''(x) = -\frac{1}{x^2} \left(1 - \frac{1}{\log x}\right) + \frac{1}{x} \cdot \frac{1}{x (\log x)^2} = -\frac{1}{x^2} + \frac{1}{x^2 \log x} + \frac{1}{x^2 (\log x)^2} \] Simplify: \[ g''(x) = -\frac{1}{x^2} + \frac{1}{x^2 \log x} + \frac{1}{x^2 (\log x)^2} \] Since \(x \in (0,1)\), \(\log x < 0\), so \(\log x\) is negative. - The dominant negative term is \(-\frac{1}{x^2}\). - The other terms are positive (since numerator and denominator are positive). **Behavior near 0:** - As \(x \to 0^+\), \(\log x \to -\infty\), so: - \(\frac{1}{\log x} \to 0^-\), - \(\frac{1}{(\log x)^2} \to 0^+\), - \(g''(x) \to -\frac{1}{x^2}\), which is negative and dominates. **Conclusion:** \(g''(x) \leq 0\) across the domain, so \(\log f(x)\) is concave. **Therefore,** \(f(x) = -x \log x\) **is log-concave** on \((0,1)\). --- ### 4. \(f(x) = \frac{x}{1 + x}\) for \(x > 0\) **Step 1:** Compute \(\log f(x)\): \[ \log f(x) = \log x - \log(1 + x) \] **Step 2:** Derivatives: \[ (\log f(x))' = \frac{1}{x} - \frac{1}{1 + x} \] \[ (\log f(x))'' = -\frac{1}{x^2} + \frac{1}{(1 + x)^2} \] **Step 3:** Analyze the sign of the second derivative: \[ (\log f(x))'' = \frac{1}{(1 + x)^2} - \frac{1}{x^2} \] - For \(x > 0\), compare: \[ \frac{1}{(1 + x)^2} \quad \text{and} \quad \frac{1}{x^2} \] - Since \(x > 0\), - \(\frac{1}{x^2} > \frac{1}{(1 + x)^2}\), - Therefore, \[ (\log f(x))'' < 0 \] **Conclusion:** \(\log f(x)\) is concave, so \(f(x)\) is **log-concave**. --- ### 5. \(f(x) = \log(1 + e^{-x})\) on \((-\infty, \infty)\) **Step 1:** \(\log f(x) = \log (\log (1 + e^{-x}))\). - The inner function: \(g(x) = \log(1 + e^{-x})\) **Step 2:** Derivatives of \(g(x)\): \[ g'(x) = \frac{-e^{-x}}{1 + e^{-x}} \] \[ g''(x) = \frac{e^{-x}}{1 + e^{-x}} - \frac{e^{-2x}}{(1 + e^{-x})^2} \] Rewrite: \[ g''(x) = \frac{e^{-x}(1 + e^{-x}) - e^{-2x}}{(1 + e^{-x})^2} = \frac{e^{-x} + e^{-2x} - e^{-2x}}{(1 + e^{-x})^2} = \frac{e^{-x}}{(1 + e^{-x})^2} > 0 \] Thus, \(g(x)\) is **convex**. Now, \(\log f(x) = \log g(x)\): - Since \(g(x)\) is convex and positive, \(\log g(x)\) is **not necessarily concave**. - In fact, the composition of a concave \(\log\) with a convex \(g(x)\) usually yields **neither convex nor concave** in general, but in particular, it can be **log-convex**. **Conclusion:** \(\log f(x)\) is **not concave**, so \(f(x)\) is **not log-concave**. --- ## **Final Summary:** - **Function 1:** Not log-concave (due to non-smoothness at \(x=2\)) - **Function 2:** Log-concave - **Function 3:** Log-concave - **Function 4:** Log-concave - **Function 5:** Not log-concave **Total number of log-concave functions:** **3** --- ### **Final answer:** \[ \boxed{3} \] **Number of log-concave functions: 3**