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);
p
r
o
o
f
:
proof:
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);
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