题目内容
方程2x-
=0的解所在区间一定是( )
| 1 |
| x |
| A.(0.4,0.5) | B.(0.5,0.6) | C.(0.6,0.7) | D.(0.7,0.8) |
根据零点的判定定理知:
令f(x)=2x-
,
则:f(0.5)=20.5-
=
-2<0;
f(0.6)=20.6-
=
-
=
-
,
因为:8-(
)5=
=
<0,
所以:f(0.6)<0;
因为:f(0.7)=20.7-
=
-
=
-
,
128-(
)10=
=
=
>0,
所以:f(0.7)>0;
所以方程2x-
=0的解所在区间一定是(0.6,0.7).
故选C.
令f(x)=2x-
| 1 |
| x |
则:f(0.5)=20.5-
| 1 |
| 0.5 |
| 2 |
f(0.6)=20.6-
| 1 |
| 0.6 |
| 5 | 8 |
| 5 |
| 3 |
| 5 | 8 |
| 5 | (
| ||
因为:8-(
| 5 |
| 3 |
| 8×243-3 125 |
| 243 |
| -1181 |
| 243 |
所以:f(0.6)<0;
因为:f(0.7)=20.7-
| 1 |
| 0.7 |
| 10 | 128 |
| 10 |
| 7 |
| 10 | 128 |
| 10 | (
| ||
128-(
| 10 |
| 7 |
| 128×710-1010 |
| 710 |
| 36156831872-10000000000 |
| 710 |
| 26156831872 |
| 710 |
所以:f(0.7)>0;
所以方程2x-
| 1 |
| x |
故选C.
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