Question

求fx＝（x²+mx）e（x次方）且m∈R的极值

Answer 1

求导,有fx′＝﹙2x＋m﹚e∧x＋﹙x²＋mx﹚e∧x＝e∧x﹙x²＋﹙m＋2﹚x＋m﹚＝0

∴x²＋﹙m＋2﹚x＋m＝0∴⊿＝﹙m＋2﹚²－4m＝m²＋4＞0

解得x1＝﹣﹙m＋2﹚＋√﹙m²＋4﹚÷﹙﹣2﹚x2＝－﹙m＋2﹚－√﹙m²＋4﹚÷﹙﹣2﹚