Ta có: f(x) = 3 => y = 3
Thay vào ta có:
\(\left|3x-1\right|-2\) = 3
=> \(\Rightarrow\left|3x-1\right|=3+2=5\)
+) 3x - 1 = 5
=> 3x = 5 + 1 = 6
=> x = \(\frac{6}{3}=2\)
+) 3x - 1 = -5
=> 3x = -5 + 1 = -4
=> x = \(\frac{-4}{3}\)
Vậy x = 2 hoặc x = \(\frac{-4}{3}\)
Ta có: \(y=f\left(x\right)=\left|3x-1\right|-2\)
Khi \(f\left(x\right)=3\) thì \(3=\left|3x-1\right|-2\)
\(\Rightarrow\left|3x-1\right|=5\)
\(\Rightarrow3x-1=\pm5\)
+) \(3x-1=5\Rightarrow x=2\)
+) \(3x-1=-5\Rightarrow x=\frac{-4}{3}\)
Vậy \(x\in\left\{2;\frac{-4}{3}\right\}\)