1. **State the problem:** Find the absolute maximum and absolute minimum values of the function $$f(x) = x^{-2} \ln(x)$$ on the interval $$\left[\frac{1}{2}, 5\right]$$. 2. **Find the derivative:** To find critical points, compute $$f'(x)$$. $$f(x) = \frac{\ln(x)}{x^2}$$ Using the quotient rule or product rule with $$f(x) = \ln(x) \cdot x^{-2}$$: $$f'(x) = \frac{1}{x} \cdot x^{-2} + \ln(x) \cdot (-2x^{-3}) = x^{-3} - 2x^{-3} \ln(x) = \frac{1 - 2\ln(x)}{x^3}$$ 3. **Find critical points:** Set $$f'(x) = 0$$: $$\frac{1 - 2\ln(x)}{x^3} = 0 \implies 1 - 2\ln(x) = 0 \implies 2\ln(x) = 1 \implies \ln(x) = \frac{1}{2}$$ 4. **Solve for $$x$$:** $$x = e^{\frac{1}{2}} = \sqrt{e} \approx 1.6487$$ 5. **Evaluate $$f(x)$$ at critical points and endpoints:** - At $$x = \frac{1}{2}$$: $$f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^{-2} \ln\left(\frac{1}{2}\right) = 4 \cdot (-\ln 2) = -4 \ln 2 \approx -2.7726$$ - At $$x = \sqrt{e}$$: $$f(\sqrt{e}) = (\sqrt{e})^{-2} \ln(\sqrt{e}) = e^{-1} \cdot \frac{1}{2} = \frac{1}{2e} \approx 0.1839$$ - At $$x = 5$$: $$f(5) = 5^{-2} \ln(5) = \frac{\ln 5}{25} \approx \frac{1.6094}{25} = 0.0644$$ 6. **Determine absolute max and min:** - Absolute minimum is $$f\left(\frac{1}{2}\right) \approx -2.7726$$ - Absolute maximum is $$f(\sqrt{e}) \approx 0.1839$$ **Final answer:** - Absolute minimum value: $$-4 \ln 2$$ at $$x = \frac{1}{2}$$ - Absolute maximum value: $$\frac{1}{2e}$$ at $$x = \sqrt{e}$$