Bài 3 trang 60 SBT toán 9 tập 1

Đề bài

Cho hàm số \(y = f\left( x \right) = \dfrac{3}{4}x\). Tính

\(f\left( { - 5} \right)\); \(f\left( { - 4} \right)\); \(f\left( { - 1} \right)\); \(f\left( 0 \right)\); \(f\left( {\dfrac{1}{2}} \right)\);

\(f\left( 1 \right)\); \(f\left( 2 \right)\); \(f\left( 4 \right)\);  \(f\left( a \right)\); \(f\left( {a + 1} \right)\).  

Lời giải chi tiết

\(f\left( { - 5} \right) = \dfrac{3}{4}.\left( { - 5} \right) =  -  \dfrac{{15}}{4}\)

\(f\left( { - 4} \right) =  \dfrac{3}{4}.\left( { - 4} \right) =  - 3\)

\(f\left( { - 1} \right) = \dfrac{3}{4}.\left( { - 1} \right) =  - \dfrac{3}{4}\)

\(f\left( 0 \right) = \dfrac{3}{4}.0 = 0\)

\(\displaystyle f\left( {{1 \over 2}} \right) = \dfrac{3}{4}.\dfrac{1}{2} = \dfrac{3}{ 8}\)

\(f\left( 1 \right) = \dfrac{3}{4}.1 = \dfrac{3}{4}\)

\(f\left( 2 \right) = \dfrac{3}{4}.2 = \dfrac{6}{4} = \dfrac{3}{ 2}\)

\(f\left( 4 \right) = \dfrac{3}{4}.4 = 3\)

\(f\left( a \right) = \dfrac{3}{4}a\)

\(f\left( {a + 1} \right) = \dfrac{3}{4}.\left( {a + 1} \right) = \dfrac{{3a + 3}}{4}\) 

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