91Ó°ÊÓ

To find the critical points of the function, first calculate the derivative of the function. Given the function \[ f(x) = x^3 - \frac{1}{2}x^2 - 2x + 5 \] the derivative is:\[ f'(x) = 3x^2 - x - 2 \]

Set the derivative equal to zero and solve for \( x \):\[ 3x^2 - x - 2 = 0 \]Using the quadratic formula \( x = \frac{-b \, \boldsymbol{\textrm{±}} \, \boldsymbol{\textrm{√}}(b^2 - 4ac)}{2a} \), where \( a = 3 \), \( b = -1 \), and \( c = -2 \):\[ x = \frac{1 \, \boldsymbol{\textrm{±}} \, \boldsymbol{\textrm{√}}(1^2 - 4 \times 3 \times -2)}{6} \]\[ x = \frac{1 \, \boldsymbol{\textrm{±}} \, \boldsymbol{\textrm{√}}(1 + 24)}{6} \]\[ x = \frac{1 \, \boldsymbol{\textrm{±}} \, \boldsymbol{\textrm{√}}25}{6} \]\[ x = \frac{1 \, \boldsymbol{\textrm{±}} \, 5}{6} \]\[ x = 1 \] or \[ x = -\frac{2}{3} \] These are the critical points.

Evaluate the original function at the critical points and at the endpoints of the interval \([-2,1]\): For \( x = -2 \):\[ f(-2) = (-2)^3 - \frac{1}{2}(-2)^2 - 2(-2) + 5 \]\[ f(-2) = -8 - 2 + 4 + 5 = -1 \] For \( x = 1 \):\[ f(1) = 1^3 - \frac{1}{2}(1)^2 - 2(1) + 5 \]\[ f(1) = 1 - \frac{1}{2} - 2 + 5 = 3.5 \] For \( x = -\frac{2}{3} \):\[ f\bigg(-\frac{2}{3}\bigg) = \bigg(-\frac{2}{3}\bigg)^3 - \frac{1}{2}\bigg(-\frac{2}{3}\bigg)^2 - 2\bigg(-\frac{2}{3}\bigg) + 5 \]\[ f\bigg(-\frac{2}{3}\bigg) = -\frac{8}{27} - \frac{1}{2} \times \frac{4}{9} + \frac{4}{3} + 5 \]\[ f\bigg(-\frac{2}{3}\bigg) = -\frac{8}{27} - \frac{2}{9} + \frac{12}{9} + 5 \]Then combine them with a common denominator (27):\[ f\bigg(-\frac{2}{3}\bigg) = -\frac{8}{27} - \frac{6}{27} + \frac{36}{27} + \frac{135}{27} \]\[ f\bigg(-\frac{2}{3}\bigg) = \frac{157}{27} = 5.81 \ (approximately ) \]

Compare the values of the function obtained at the critical points and the endpoints: \( f(-2) = -1 \) \( f(1) = 3.5 \) \( f\bigg(-\frac{2}{3}\bigg) = 5.81 \)Therefore, the absolute maximum value is 5.81 at \( x = -\frac{2}{3} \), and the absolute minimum value is -1 at \( x = -2 \).