1. **State the problem:** We want to analyze the function $$f(x) = \frac{1}{e^x + 1}$$ using its first and second derivatives to understand its behavior and sketch its graph. 2. **Find the first derivative:** Use the quotient rule or rewrite as $$f(x) = (e^x + 1)^{-1}$$ and apply the chain rule. $$f'(x) = -1 \cdot (e^x + 1)^{-2} \cdot e^x = -\frac{e^x}{(e^x + 1)^2}$$ 3. **Analyze the first derivative:** Since $$e^x > 0$$ for all real $$x$$, the numerator $$e^x$$ is positive and the denominator $$(e^x + 1)^2$$ is always positive. The negative sign means $$f'(x) < 0$$ for all $$x$$. **Conclusion:** The function is strictly decreasing everywhere. 4. **Find the second derivative:** Differentiate $$f'(x) = -\frac{e^x}{(e^x + 1)^2}$$ using the quotient rule: Let $$u = e^x$$ and $$v = (e^x + 1)^2$$. $$f''(x) = - \frac{u'v - uv'}{v^2} = - \frac{e^x (e^x + 1)^2 - e^x \cdot 2 (e^x + 1) e^x}{(e^x + 1)^4}$$ Simplify numerator: $$e^x (e^x + 1)^2 - 2 e^{2x} (e^x + 1) = e^x (e^x + 1) [ (e^x + 1) - 2 e^x ] = e^x (e^x + 1)(1 - e^x)$$ So, $$f''(x) = - \frac{e^x (e^x + 1)(1 - e^x)}{(e^x + 1)^4} = - \frac{e^x (1 - e^x)}{(e^x + 1)^3}$$ 5. **Analyze the second derivative:** - The denominator is positive. - The numerator sign depends on $$e^x (1 - e^x)$$. Since $$e^x > 0$$: - If $$e^x < 1$$ (i.e., $$x < 0$$), then $$1 - e^x > 0$$, so numerator positive, but with the leading negative sign, $$f''(x) < 0$$. - If $$e^x > 1$$ (i.e., $$x > 0$$), then $$1 - e^x < 0$$, numerator negative, so $$f''(x) > 0$$. 6. **Concavity and inflection point:** - $$f''(x) < 0$$ for $$x < 0$$: function is concave down. - $$f''(x) > 0$$ for $$x > 0$$: function is concave up. - At $$x=0$$, $$f''(0) = 0$$, so there is an inflection point. 7. **Limits and behavior:** - As $$x \to -\infty$$, $$e^x \to 0$$, so $$f(x) \to \frac{1}{0 + 1} = 1$$. - As $$x \to \infty$$, $$e^x \to \infty$$, so $$f(x) \to \frac{1}{\infty} = 0$$. 8. **Summary:** - $$f(x)$$ is strictly decreasing from 1 to 0. - Concave down for $$x < 0$$, concave up for $$x > 0$$. - Inflection point at $$x=0$$. **Final answer:** The function $$f(x) = \frac{1}{e^x + 1}$$ decreases smoothly from 1 to 0 with an inflection point at 0 where concavity changes from down to up.