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定义Mnx=x(x+1)(x+2)…(x+n−1)(x∈R,n∈N*),如M4−4=(−4)×(−3)×(−2)×(−1)=24.对于函数f(x)=M3x−1,则函数f(x)的解析式是:,且f(x)的单调递减区间是(−33,33)(−33,33)(写成开
题目详情
定义
=x(x+1)(x+2)…(x+n−1)(x∈R,n∈N*),如
=(−4)×(−3)×(−2)×(−1)=24.对于函数f(x)=
,则函数f(x)的解析式是:______,且f(x)的单调递减区间是
=x(x+1)(x+2)…(x+n−1)(x∈R,n∈N*),如
=(−4)×(−3)×(−2)×(−1)=24.对于函数f(x)=
,则函数f(x)的解析式是:______,且f(x)的单调递减区间是
M M
=(−4)×(−3)×(−2)×(−1)=24.对于函数f(x)=
,则函数f(x)的解析式是:______,且f(x)的单调递减区间是
=(−4)×(−3)×(−2)×(−1)=24.对于函数f(x)=
,则函数f(x)的解析式是:______,且f(x)的单调递减区间是
=(−4)×(−3)×(−2)×(−1)=24.对于函数f(x)=
,则函数f(x)的解析式是:______,且f(x)的单调递减区间是
=(−4)×(−3)×(−2)×(−1)=24.对于函数f(x)=
,则函数f(x)的解析式是:______,且f(x)的单调递减区间是
M M
,则函数f(x)的解析式是:______,且f(x)的单调递减区间是
M M
,
)
3 3 3 3
3 3 3 3
,
)
3 3 3 3
3 3 3 3
M | n x |
M | 4 −4 |
M | 3 x−1 |
(−
,
)
| ||
3 |
| ||
3 |
(−
,
)
(写成开区间或闭区间都给全分).
| ||
3 |
| ||
3 |
M | n x |
M | 4 −4 |
M | 3 x−1 |
(−
,
)
| ||
3 |
| ||
3 |
(−
,
)
(写成开区间或闭区间都给全分).
| ||
3 |
| ||
3 |
M | n x |
n
x
n
x
n
x
N*),如M | 4 −4 |
M | 3 x−1 |
(−
,
)
| ||
3 |
| ||
3 |
(−
,
)
(写成开区间或闭区间都给全分).N*),如
| ||
3 |
| ||
3 |
M | 4 −4 |
M | 3 x−1 |
(−
,
)
| ||
3 |
| ||
3 |
(−
,
)
(写成开区间或闭区间都给全分).*),如
| ||
3 |
| ||
3 |
M | 4 −4 |
M | 3 x−1 |
(−
,
)
| ||
3 |
| ||
3 |
(−
,
)
(写成开区间或闭区间都给全分).
| ||
3 |
| ||
3 |
M | 4 −4 |
M | 3 x−1 |
(−
,
)
| ||
3 |
| ||
3 |
(−
,
)
(写成开区间或闭区间都给全分).
| ||
3 |
| ||
3 |
M | 4 −4 |
4
−4
4
−4
4
−4
f(x)=M | 3 x−1 |
(−
,
)
| ||
3 |
| ||
3 |
(−
,
)
(写成开区间或闭区间都给全分).
| ||
3 |
| ||
3 |
M | 3 x−1 |
3
x−1
3
x−1
3
x−1
(−
,
)
(−
| ||
3 |
| ||
3 |
| ||
3 |
| ||
3 |
| ||
3 |
3 |
3 |
3 |
| ||
3 |
3 |
3 |
3 |
(−
,
)
(−
| ||
3 |
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3 |
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3 |
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3 |
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3 |
3 |
3 |
3 |
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3 |
3 |
3 |
3 |
▼优质解答
答案和解析
∵f(x)=(x-1)•x•(x+1)=x33-x,
又由f′(x)=3x22-1<0,得−
<x<
即f(x)的单调减区间为(−
,
).
故答案为:f(x)=x3-x (−
,
) −
3 3 33 3 3<x<
3 3 33 3 3
即f(x)的单调减区间为(−
,
).
故答案为:f(x)=x3-x (−
,
) (−
3 3 33 3 3,
3 3 33 3 3).
故答案为:f(x)=x33-x (−
,
) (−
3 3 33 3 3,
3 3 33 3 3)
又由f′(x)=3x22-1<0,得−
| ||
3 |
| ||
3 |
即f(x)的单调减区间为(−
| ||
3 |
| ||
3 |
故答案为:f(x)=x3-x (−
| ||
3 |
| ||
3 |
| ||
3 |
3 |
3 |
3 |
| ||
3 |
3 |
3 |
3 |
即f(x)的单调减区间为(−
| ||
3 |
| ||
3 |
故答案为:f(x)=x3-x (−
| ||
3 |
| ||
3 |
| ||
3 |
3 |
3 |
3 |
| ||
3 |
3 |
3 |
3 |
故答案为:f(x)=x33-x (−
| ||
3 |
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3 |
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3 |
3 |
3 |
3 |
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3 |
3 |
3 |
3 |
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