Ta có: \(y = f\left( x \right) = \dfrac{{12}}{x}\)
a) \(f\left( 5 \right) =\dfrac{{12}}{5} = 2,4\)
\( f\left( { - 3} \right) = \dfrac{{12}}{{ - 3}} = - 4\)
b)
\(\eqalign{
& f\left( { - 6} \right) = {{12} \over { - 6}} = - 2 \cr
& f\left( { - 4} \right) = {{12} \over { - 4}} = - 3 \cr
& f\left( { - 3} \right) = {{12} \over { - 3}} = - 4 \cr
& f\left( 2 \right) = {{12} \over 2} = 6 \cr
& f\left( 5 \right) = {{12} \over 5} = 2,4 \cr
& f\left( 6 \right) = {{12} \over 6} = 2 \cr
& f\left( {12} \right) = {{12} \over {12}} = 1 \cr} \)
Ta được bảng sau:
|
x
|
-6
|
-4
|
-3
|
2
|
5
|
6
|
12
|
|
\(f\left( x \right) = \dfrac{{12}}{x}\)
|
-2
|
-3
|
-4
|
6
|
2,4
|
2
|
1
|
