1. Find the derivative: \[ f'(x) = \frac{1}{2} - \frac{2}{x^2}. \] 2. Critical points: Set \( f'(x) = 0 \): \[ \frac{1}{2} - \frac{2}{x^2} = 0 \quad \Rightarrow \quad \frac{2}{x^2} = \frac{1}{2} \quad \Rightarrow \quad x^2 = 4 \quad \Rightarrow \quad x = 2. \] (Only \( x = 2 \) is in \( [1, 2] \).)
3. Evaluate \( f(x) \) at endpoints and critical point: - At \( x = 1 \): \[ f(1) = \frac{1}{2} + \frac{2}{1} = \frac{1}{2} + 2 = 2.5. \] - At \( x = 2 \): \[ f(2) = \frac{2}{2} + \frac{2}{2} = 1 + 1 = 2. \]
4. Conclusion: The absolute maximum value is \( 2.5 \) at \( x = 1 \), and the absolute minimum value is \( 2 \) at \( x = 2 \).
Final Answer: \[ {Absolute maximum: } 2.5 \, ({at } x = 1), \quad {Absolute minimum: } 2 \, ({at } x = 2). \]
Determine whether each of the following relations are reflexive, symmetric, and transitive.
Show that the relation R in the set R of real numbers, defined as
R = {(a, b): a ≤ b2 } is neither reflexive nor symmetric nor transitive.
Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.