已知函数f(x)=e的x次方-ax,a∈R,e是自然对数的底数(1)若函数在x=2处取得极值?
如题所述
详细过程如图请参考
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/aa18972bd40735faf67db9c58e510fb30f240838?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/b812c8fcc3cec3fd05ac5009c688d43f87942738?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
温馨提示:答案为网友推荐,仅供参考
相似回答
已知函数f(x)=e^x-ax-1(a>0..e为自然对数的底数).已知函数f...答:f'(x)=e^x-a,x=㏑a取最小值,f(a)=a-a㏑a-1.2)欲使f(a)≥0,f'(a)=-㏑a,所以f(a)在a=1处取最大值0,故a只能取1.(3)当a=1时,f(x)=e^x-x-1.由(2)知,对于任意x∈R,f(x)≥0,令x=t-1,∴f(t-1)=e^(t-1)-(t-1)-1=e^(t-1)-t≥0.∴t≤e...