1. **State the problem:** Find the absolute extreme values of the function $$f(x) = \ln(x + 2) + \frac{1}{x}$$ on the interval $$[1, 10]$$. 2. **Find the derivative:** To locate critical points, compute $$f'(x)$$. $$f'(x) = \frac{1}{x+2} - \frac{1}{x^2}$$ 3. **Set the derivative equal to zero to find critical points:** $$\frac{1}{x+2} - \frac{1}{x^2} = 0$$ Multiply both sides by $$x^2(x+2)$$ to clear denominators: $$x^2 - (x+2) = 0$$ Simplify: $$x^2 - x - 2 = 0$$ 4. **Solve the quadratic equation:** $$x^2 - x - 2 = 0$$ Using the quadratic formula: $$x = \frac{1 \pm \sqrt{1 + 8}}{2} = \frac{1 \pm 3}{2}$$ So, $$x = 2 \quad \text{or} \quad x = -1$$ Since the domain is $$[1, 10]$$, discard $$x = -1$$. 5. **Evaluate the function at critical points and endpoints:** - At $$x=1$$: $$f(1) = \ln(3) + 1 = \ln(3) + 1 \approx 1.0986 + 1 = 2.0986$$ - At $$x=2$$: $$f(2) = \ln(4) + \frac{1}{2} = \ln(4) + 0.5 \approx 1.3863 + 0.5 = 1.8863$$ - At $$x=10$$: $$f(10) = \ln(12) + \frac{1}{10} = \ln(12) + 0.1 \approx 2.4849 + 0.1 = 2.5849$$ 6. **Determine absolute extrema:** - The absolute minimum value is $$f(2) \approx 1.8863$$. - The absolute maximum value is $$f(10) \approx 2.5849$$. **Final answer:** - Absolute minimum at $$x=2$$ with value $$\approx 1.8863$$. - Absolute maximum at $$x=10$$ with value $$\approx 2.5849$$.