(a) The graph of \(y = |4x - 2|\) is a V-shaped graph. The vertex of the graph is at \(x = \frac{1}{2}\), \(y = 0\). The graph is symmetrical about the vertical line \(x = \frac{1}{2}\) and extends into the second quadrant.
(b) To solve \(1 + 3x < |4x - 2|\), consider two cases:
Case 1: \(4x - 2 \geq 0\) (i.e., \(x \geq \frac{1}{2}\))
Then \(|4x - 2| = 4x - 2\).
Solving \(1 + 3x < 4x - 2\) gives:
\(1 + 3x < 4x - 2\)
\(1 + 3x - 4x < -2\)
\(1 - x < -2\)
\(-x < -3\)
\(x > 3\)
Case 2: \(4x - 2 < 0\) (i.e., \(x < \frac{1}{2}\))
Then \(|4x - 2| = -(4x - 2) = -4x + 2\).
Solving \(1 + 3x < -4x + 2\) gives:
\(1 + 3x < -4x + 2\)
\(1 + 3x + 4x < 2\)
\(1 + 7x < 2\)
\(7x < 1\)
\(x < \frac{1}{2}\)
Thus, the solution is \(x < \frac{1}{2}\) or \(x > 3\).
