Solve |linear| = linear (non-modulus)

3. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of the graph with equation \(y = 2 | x | - 5\).
The graph intersects the positive \(x\)-axis at the point \(P\) and the negative \(y\)-axis at the point \(Q\).

1. State the coordinates of \(P\) and the coordinates of \(Q\).
2. Solve the equation $$2 | x | - 5 = 3 - x$$

1. The functions \(f\) and \(g\) are defined by

$$\begin{aligned} & \mathrm { f } : x \rightarrow | 2 x - 5 | , \quad x \in \mathbb { R } , \\ & \mathrm {~g} : x \rightarrow \ln ( x + 3 ) , \quad x \in \mathbb { R } , \quad x > - 3 \end{aligned}$$

1. State the range of f .
2. Evaluate fg(-2).
3. Solve the equation $$\operatorname { fg } ( x ) = 3$$ giving your answers in exact form.
4. Show that the equation $$\mathrm { f } ( x ) = \mathrm { g } ( x )$$ has a root, \(\alpha\), in the interval [3,4].
5. Use the iterative formula $$x _ { n + 1 } = \frac { 1 } { 2 } \left[ 5 + \ln \left( x _ { n } + 3 \right) \right]$$ with \(x _ { 0 } = 3\), to find \(x _ { 1 } , x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving your answers to 4 significant figures.
6. Show that your answer for \(x _ { 4 }\) is the value of \(\alpha\) correct to 4 significant figures.

1 Solve the equation \(| 3 x - 2 | = x\).

11 Solve the equation \(| 3 x - 2 | = x\).

1. In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of the graph with equation \(y = | 3 - 2 x |\) Solve $$| 3 - 2 x | = 7 + x$$

Solve the equation \(3 + 2x = |7 - 4x|\). [3]