分析:(1)當(dāng)a=0時(shí),
f′(x)=.故f(x)的單調(diào)增區(qū)間是(0,+∞);單調(diào)減區(qū)間是(-1,0).當(dāng)a>0時(shí),令f'(x)=0,得x
1=0,或
x2=-1.當(dāng)0<a<1時(shí),列表討論f(x)與f'(x)的情況能求出f(x)的單調(diào)區(qū)間.
(3)
+
+
+…+
=3
n-(
+++…+)<3
n-
(n∈N
*).
解答:
解:(1)∵f(x)=x-
ax
2-ln(1+x),
∴f′(x)=1-ax-
=
,x>-1,
:①當(dāng)a=0時(shí),
f′(x)=.
故f(x)的單調(diào)增區(qū)間是(0,+∞);單調(diào)減區(qū)間是(-1,0).
②當(dāng)a>0時(shí),令f'(x)=0,得x
1=0,或x
2=
-1.
當(dāng)0<a<1時(shí),f(x)與f'(x)的情況如下:
當(dāng)a>0時(shí),令f′(x)=0,得x
1=0,或x
2=
-1.
當(dāng)0<a<1時(shí),f(x)與f'(x)的情況如下:
x | (-1,x1) | x1 | (x1,x2) | x2 | (x2,+∞) |
f′(x) | - | 0 | + | 0 | - |
f(x) | ↘ | f(x1) | ↗ | f(x2) | ↘ |
所以,f(x)的單調(diào)增區(qū)間是(0,
-1);單調(diào)減區(qū)間是(-1,0)和(
-1,+∞).
當(dāng)a=1時(shí),f(x)的單調(diào)減區(qū)間是(-1,+∞).
當(dāng)a>1時(shí),-1<x
2<0,f(x)與f'(x)的情況如下:
x | (-1,x2) | x2 | (x2,x1) | x1 | (x1,+∞) |
f′(x) | - | 0 | + | 0 | - |
f(x) | ↘ | f(x2) | ↗ | f(x1) | ↘ |
所以,f(x)的單調(diào)增區(qū)間是(
-1,0);單調(diào)減區(qū)間是(-1,
-1)和(0,+∞).
③當(dāng)a<0時(shí),f(x)的單調(diào)增區(qū)間是(0,+∞);單調(diào)減區(qū)間是(-1,0).
綜上,當(dāng)a≤0時(shí),f(x)的增區(qū)間是(0,+∞),減區(qū)間是(-1,0);
當(dāng)0<a<1時(shí),f(x)的增區(qū)間是(0,
-1),減區(qū)間是(-1,0)和(
-1,+∞);
當(dāng)a=1時(shí),f(x)的減區(qū)間是(-1,+∞);
當(dāng)a>1時(shí),f(x)的增區(qū)間是(
-1,0);減區(qū)間是(-1,
-1)和(0,+∞).
(3)證明:
+
+
+…+
=(1+
)+(1+
)+(1+
)+…+(1+
)
=3
n-(
+++…+)
<3
n-
(n∈N
*).