放缩不等式推导

放缩不等式推导

1 ) a x > x + 1 ( 1 < a ≤ e , x < 0 ; a ≥ e , x > 0 ) ; 1)\\ a^x>x+1\\left(1<a\\leq e,x<0;a\\geq e,x>0\\right); 1) ax>x+1(1<a≤e,x<0;a≥e,x>0);
$proof:$
f 01 ( x ) = a x − ( x + 1 ) ⇒ f 01 ′ ( x ) = a x ln ⁡ a − 1 f_{01}\\left(x\\right)=a^{x}-\\left(x+1\\right)\\Rightarrow f_{01}^{'}\\left(x\\right) = a ^{x} \\ln a-1 f01​(x)=ax−(x+1)⇒f01′​(x)=axlna−1
1 < a ≤ e , x < 0 ⇒ 0 < a x < 1 , 0 < ln ⁡ a ≤ 1 ⇒ f 01 ( x ) > f 01 ( 0 ) = 1 − 1 = 0 ⇒ a x > x + 1 ( 1 < a ≤ e , x < 0 ) ; 1<a\\leq e,x<0\\\\\\Rightarrow0<a^{x}<1,0<\\ln a\\leq1\\\\\\Rightarrow f_{01}\\left(x\\right)>f_{01}\\left(0\\right)=1-1=0\\\\\\Rightarrow a^{x}>x+1\\left(1<a\\leq e,x<0\\right); 1<a≤e,x<0⇒0<ax<1,0<lna≤1⇒f01​(x)>f01​(0)=1−1=0⇒ax>x+1(1<a≤e,x<0);