Answer
$${\text{no minimum; max}} = 5{\text{ at }}x = - 1$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = 3 - 4x - 2{x^2};\,\,\left( { - \infty , + \infty } \right) \cr
& {\text{The function is a polynomial }} \cr
& {\text{Since }}f(x){\text{ has even }}\,{\text{degree and the leading coefficient is negative}} \cr
& f(x) \to - \infty {\text{ as }}x \to \pm \infty .{\text{ Thus}}{\text{, there is an absolute maximum but}} \cr
& {\text{not absolute minimum}}{\text{. From }}\,{\text{theorem}}\,{\text{ 3}}{\text{.4}}{\text{.3}}{\text{, }}\,{\text{the absolute maximum}} \cr
& {\text{must occur at a critical point }}f. \cr
& {\text{Since }}f{\text{ is differentiable everywhere}}{\text{, we can find all critical points }} \cr
& {\text{by solving the equation }}f'(x) = 0.{\text{ This equation}}\,{\text{is}} \cr
& f'\left( x \right) = - 4 - 4x \cr
& {\text{Let }}f'\left( x \right) = 0 \cr
& - 4 - 4x = 0 \cr
& x = - 1 \cr
& {\text{Evaluating }}f{\text{ at the critical point}} \cr
& f\left( { - 1} \right) = 3 - 4\left( { - 1} \right) - 2{\left( { - 1} \right)^2} \cr
& f\left( { - 1} \right) = 3 + 4 - 2 \cr
& f\left( { - 1} \right) = 5 \cr
& {\text{Therefore}}{\text{, }}f{\text{ has an absolute maximum of }}5{\text{ at }}x = - 1 \cr} $$
