The problem provides a function $f(x) = \frac{1}{4} - (x - \frac{3}{2})^2$ and asks to: 1) Develop, reduce and order $f(x)$. 2) Show that $f(x) = (x-1)(-x+2)$. 3) Solve equations: a) $\frac{2x+6}{2} = 3$ b) $|-3x+4| = |4+\frac{7}{2}|$ c) $\frac{x-3}{2} - \frac{x+1}{3} = 0$ e) $f(x) = 0$ and $f(x) = -2$ g) $(x - \frac{3}{2})^2 = (3x - \frac{1}{2})^2$

1. Problem Description

The problem provides a function $f(x) = \frac{1}{4} - (x - \frac{3}{2})^2$ and asks to:

1) Develop, reduce and order $f(x)$.

2) Show that $f(x) = (x-1)(-x+2)$.

3) Solve equations:

a) $\frac{2x+6}{2} = 3$

b) $|-3x+4| = |4+\frac{7}{2}|$

c) $\frac{x-3}{2} - \frac{x+1}{3} = 0$

e) $f(x) = 0$ and $f(x) = -2$

g) $(x - \frac{3}{2})^2 = (3x - \frac{1}{2})^2$

2. Solution Steps

1) Develop, reduce and order $f(x)$.

$f(x) = \frac{1}{4} - (x - \frac{3}{2})^2$

$f(x) = \frac{1}{4} - (x^2 - 2 \cdot x \cdot \frac{3}{2} + (\frac{3}{2})^2)$

$f(x) = \frac{1}{4} - (x^2 - 3x + \frac{9}{4})$

$f(x) = \frac{1}{4} - x^2 + 3x - \frac{9}{4}$

$f(x) = -x^2 + 3x - \frac{8}{4}$

$f(x) = -x^2 + 3x - 2$

2) Show that $f(x) = (x-1)(-x+2)$.

$(x-1)(-x+2) = -x^2 + 2x + x - 2 = -x^2 + 3x - 2 = f(x)$

3) Solve equations:

a) $\frac{2x+6}{2} = 3$

$2x + 6 = 6$

$2x = 0$

$x = 0$

b) $|-3x+4| = |4+\frac{7}{2}|$

$|-3x+4| = |\frac{8+7}{2}|$

$|-3x+4| = |\frac{15}{2}|$

$-3x+4 = \frac{15}{2}$ or $-3x+4 = -\frac{15}{2}$

$-3x = \frac{15}{2} - 4 = \frac{15-8}{2} = \frac{7}{2}$ or $-3x = -\frac{15}{2} - 4 = \frac{-15-8}{2} = -\frac{23}{2}$

$x = -\frac{7}{6}$ or $x = \frac{23}{6}$

c) $\frac{x-3}{2} - \frac{x+1}{3} = 0$

$\frac{3(x-3) - 2(x+1)}{6} = 0$

$3x - 9 - 2x - 2 = 0$

$x - 11 = 0$

$x = 11$

e) $f(x) = 0$

$-x^2 + 3x - 2 = 0$

$(x-1)(-x+2) = 0$

$x-1 = 0$ or $-x+2 = 0$

$x = 1$ or $x = 2$

$f(x) = -2$

$-x^2 + 3x - 2 = -2$

$-x^2 + 3x = 0$

$x(-x+3) = 0$

$x = 0$ or $-x+3 = 0$

$x = 0$ or $x = 3$

g) $(x - \frac{3}{2})^2 = (3x - \frac{1}{2})^2$

$x - \frac{3}{2} = 3x - \frac{1}{2}$ or $x - \frac{3}{2} = -(3x - \frac{1}{2})$

$-2x = \frac{3}{2} - \frac{1}{2} = \frac{2}{2} = 1$ or $x - \frac{3}{2} = -3x + \frac{1}{2}$

$x = -\frac{1}{2}$ or $4x = \frac{3}{2} + \frac{1}{2} = \frac{4}{2} = 2$

$x = -\frac{1}{2}$ or $x = \frac{1}{2}$

3. Final Answer

1) $f(x) = -x^2 + 3x - 2$

2) $f(x) = (x-1)(-x+2)$

3) a) $x = 0$

b) $x = -\frac{7}{6}$ or $x = \frac{23}{6}$

c) $x = 11$

e) $x = 1$ or $x = 2$, $x = 0$ or $x = 3$

g) $x = -\frac{1}{2}$ or $x = \frac{1}{2}$