To confirm this critical point is a local minimum we use the first derivative test. For {{MATH_21}}, {{MATH_22}} is {{CMD_23}}{{CMD_24}}positive{{CMD_25}}{{CMD_26}}negative{{CMD_27}}{{CMD_28}}, and for {{MATH_29}}, {{MATH_30}} is {{CMD_31}}{{CMD_32}}positive{{CMD_33}}{{CMD_34}}negative{{CMD_35}}{{CMD_36}}. Then by the first derivative test, {{MATH_37}} has a local minimum. Now since {{MATH_38}} has only one critical point on its domain, and it has a local minimum there, then that local minimum is also the absolute minimum of {{MATH_39}}.