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1.
y=f(-1)=3*(-1)-2=-5
y=f(0)=3*0-2=-2
y=f(-2)=3*(-2)-2=-8
y=f(3)=3*3-2=7
Câu 2,3a làm tương tự,chỉ việc thay f(x) thôi.
3b
Khi y=5 =>5=5-2*x=>2*x=0=> x=0
Khi y=3=>3=5-2*x=>2*x=2=>x=1
Khi y=-1=>-1=5-2*x=>2*x=6=>x=3
f(-1)=3.1-2=3-2=1
f(0)=3.0-2=0-2=-2
f(-2)=3.(-2)-2=-6-2=-8
f(3)=3.3-2=9-2=7
\(a,f\left(1\right)=3\cdot1^2+1+1=5\\ f\left(-\dfrac{1}{3}\right)=3\cdot\left(-\dfrac{1}{3}\right)^2-\dfrac{1}{3}+1=\dfrac{1}{3}-\dfrac{1}{3}+1=1\\ f\left(\dfrac{2}{3}\right)=3\cdot\left(\dfrac{2}{3}\right)^2-\dfrac{2}{3}+1=\dfrac{4}{3}-\dfrac{2}{3}+1=\dfrac{5}{3}\\ f\left(-2\right)=3\cdot\left(-2\right)^2-2+1=11\\ f\left(-\dfrac{4}{3}\right)=3\cdot\left(-\dfrac{4}{3}\right)^2-\dfrac{4}{3}+1=\dfrac{16}{3}-\dfrac{4}{3}+1=5\)
\(b,f\left(\dfrac{2}{3}\right)=\left|2\cdot\dfrac{2}{3}-9\right|-3=\dfrac{23}{3}-3=\dfrac{14}{3}\\ f\left(-\dfrac{5}{4}\right)=\left|2\cdot\left(-\dfrac{5}{4}\right)-9\right|-3=\dfrac{23}{2}-3=\dfrac{17}{2}\\ f\left(-5\right)=\left|2\left(-5\right)-9\right|-3=19-3=16\\ f\left(4\right)=\left|2\cdot4-9\right|-3=1-3=-2\\ f\left(-\dfrac{3}{8}\right)=\left|2\cdot\left(-\dfrac{3}{8}\right)-9\right|-3=\dfrac{39}{4}-3=\dfrac{27}{4}\)
\(c,x=0\Rightarrow y=2\cdot0^2-7=-7\\ x=-3\Rightarrow y=2\cdot\left(-3\right)^2-7=11\\ x=-\dfrac{1}{2}\Rightarrow y=2\cdot\left(-\dfrac{1}{2}\right)^2-7=\dfrac{-13}{2}\\ x=\dfrac{2}{3}\Rightarrow y=2\cdot\left(\dfrac{2}{3}\right)^2-7=-\dfrac{55}{9}\)
a ) Ta có : f(2) = 5
\(\Leftrightarrow\hept{\begin{cases}f\left(x\right)=f\left(2\right)\\\text{ax}-3=5\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=2\\a.2-3=5\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=2\\a=4\end{cases}}\)
Vậy a = 4
b ) Ta có : f(0) = 3
\(\Leftrightarrow\hept{\begin{cases}f\left(x\right)=f\left(0\right)\\\text{ax}+b=3\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\a.0+b=3\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\b=3\end{cases}}\) ( 1 )
Ta có : f ( 1 ) = 4
\(\Leftrightarrow\hept{\begin{cases}f\left(x\right)=f\left(1\right)\\\text{ax}+b=4\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=1\\a.1+b=4\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=1\\a+b=4\end{cases}}\) ( 2 )
Thay b = 3 ở ( 1 ) vào a+b=4 ở ( 2 ) ta được : a + 3 = 4
a = 1
Vậy a = 1 ; b = 3
a) Thay x=-2 vào hàm số f(x)=|3x-1|, ta được:
\(f\left(-2\right)=\left|3\cdot\left(-2\right)-1\right|=\left|-6-1\right|=7\)
Thay x=2 vào hàm số \(f\left(x\right)=\left|3x-1\right|\), ta được:
\(f\left(2\right)=\left|3\cdot2-1\right|=\left|6-1\right|=5\)
Thay \(x=-\dfrac{1}{4}\) vào hàm số \(f\left(x\right)=\left|3x-1\right|\), ta được:
\(f\left(-\dfrac{1}{4}\right)=\left|3\cdot\dfrac{-1}{4}-1\right|=\left|-\dfrac{3}{4}-\dfrac{4}{4}\right|=\dfrac{7}{4}\)
Thay \(x=\dfrac{1}{4}\) vào hàm số \(f\left(x\right)=\left|3x-1\right|\), ta được:
\(f\left(\dfrac{1}{4}\right)=\left|3\cdot\dfrac{1}{4}-1\right|=\left|\dfrac{3}{4}-1\right|=\dfrac{1}{4}\)
Vậy: f(-2)=7; f(2)=5; \(f\cdot\left(-\dfrac{1}{4}\right)=\dfrac{7}{4}\); \(f\left(\dfrac{1}{4}\right)=\dfrac{1}{4}\)
b) Để f(x)=10 thì \(\left|3x-1\right|=10\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=10\\3x-1=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=11\\3x=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{3}\\x=-3\end{matrix}\right.\)
Để f(x)=-3 thì \(\left|3x-1\right|=-3\)
mà \(\left|3x-1\right|\ge0\forall x\)
nên \(x\in\varnothing\)
Trả lời:
\(f\left(0\right)=3.0+29-2=27\)
\(f\left(4\right)=3.4+29-2=39\)
\(f\left(-3\right)=3.\left(-3\right)+29-2=18\)
\(f\left(5\right)=3.5+29-2=42\)
Học tốt
42 nha!!!!!!!!!!!!!!!!!!!!!