1. **State the problem:** We want to find the direction of change (increasing or decreasing) of the function $$f(x) = \sqrt{x} - 1 - \sqrt{x} \ln(x)$$ for $$x > 0$$ since the function involves $$\ln(x)$$ and $$\sqrt{x}$$ which are defined for positive $$x$$ only. 2. **Find the derivative:** The direction of change is determined by the sign of the derivative $$f'(x)$$. Given $$f(x) = \sqrt{x} - 1 - \sqrt{x} \ln(x)$$, rewrite $$\sqrt{x}$$ as $$x^{1/2}$$: $$f(x) = x^{1/2} - 1 - x^{1/2} \ln(x)$$ Differentiate term by term: - Derivative of $$x^{1/2}$$ is $$\frac{1}{2} x^{-1/2}$$ - Derivative of $$-1$$ is $$0$$ - Derivative of $$-x^{1/2} \ln(x)$$ uses product rule: Let $$u = x^{1/2}$$ and $$v = \ln(x)$$ $$u' = \frac{1}{2} x^{-1/2}$$ $$v' = \frac{1}{x}$$ So, $$\frac{d}{dx} (u v) = u' v + u v' = \frac{1}{2} x^{-1/2} \ln(x) + x^{1/2} \cdot \frac{1}{x} = \frac{1}{2} x^{-1/2} \ln(x) + x^{-1/2}$$ Therefore, $$f'(x) = \frac{1}{2} x^{-1/2} - \left( \frac{1}{2} x^{-1/2} \ln(x) + x^{-1/2} \right) = \frac{1}{2} x^{-1/2} - \frac{1}{2} x^{-1/2} \ln(x) - x^{-1/2}$$ 3. **Simplify the derivative:** $$f'(x) = \frac{1}{2} x^{-1/2} - \frac{1}{2} x^{-1/2} \ln(x) - x^{-1/2} = x^{-1/2} \left( \frac{1}{2} - \frac{1}{2} \ln(x) - 1 \right) = x^{-1/2} \left( -\frac{1}{2} - \frac{1}{2} \ln(x) \right)$$ Rewrite: $$f'(x) = -\frac{1}{2} x^{-1/2} (1 + \ln(x))$$ 4. **Analyze the sign of $$f'(x)$$:** - $$x^{-1/2} > 0$$ for $$x > 0$$ - The factor $$-\frac{1}{2}$$ is negative - The sign depends on $$1 + \ln(x)$$ Set $$1 + \ln(x) = 0$$ to find critical point: $$\ln(x) = -1 \implies x = e^{-1} = \frac{1}{e} \approx 0.3679$$ - For $$x < \frac{1}{e}$$, $$\ln(x) < -1$$ so $$1 + \ln(x) < 0$$, then $$f'(x) = -\frac{1}{2} x^{-1/2} (negative) = positive$$ - For $$x > \frac{1}{e}$$, $$\ln(x) > -1$$ so $$1 + \ln(x) > 0$$, then $$f'(x) = -\frac{1}{2} x^{-1/2} (positive) = negative$$ 5. **Conclusion:** - $$f(x)$$ is increasing on $$\left(0, \frac{1}{e}\right)$$ - $$f(x)$$ is decreasing on $$\left(\frac{1}{e}, \infty\right)$$ Hence, the function increases until $$x = \frac{1}{e}$$ and then decreases afterwards.