1.2. For each of the following problems, show that $$ (f \circ g)(x)=(g \circ f)(x)=x $$ $$ \left.1.2 .1 f(x)=a x+b, \quad g(x)=\frac{1}{a}(x-b) ight), a
eq 0 $$ $$ 1.2 .2 f(x)=\frac{1}{x}, g(x)=\frac{1}{x} $$ $$ 1.2 .3 f(x)=2 x, g(x)=\frac{1}{2} x $$ $$ 1.2 .4 \quad f(x)=\frac{2 x+1}{x-1}, g(x)=\frac{x+1}{x-2} x $$

Question

1.2. For each of the following problems, show that $$ (f \circ g)(x)=(g \circ f)(x)=x $$ $$ \left.1.2 .1 f(x)=a x+b, \quad g(x)=\frac{1}{a}(x-b)ight), a 
eq 0 $$ $$ 1.2 .2 f(x)=\frac{1}{x}, g(x)=\frac{1}{x} $$ $$ 1.2 .3 f(x)=2 x, g(x)=\frac{1}{2} x $$ $$ 1.2 .4 \quad f(x)=\frac{2 x+1}{x-1}, g(x)=\frac{x+1}{x-2} x $$

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To show that $$ (f \circ g)(x) = (g \circ f)(x) = x $$ for each pair of functions, we will compute the compositions $$ f(g(x)) $$ and $$ g(f(x)) $$ for each case.

### 1.2.1: $$ f(x) = ax + b $$, $$ g(x) = \frac{1}{a}(x - b) $$

1. **Calculate $$ (f \circ g)(x) $$**:
   $$
   f(g(x)) = f\left(\frac{1}{a}(x - b)\right) = a\left(\frac{1}{a}(x - b)\right) + b = x - b + b = x
   $$

2. **Calculate $$ (g \circ f)(x) $$**:
   $$
   g(f(x)) = g(ax + b) = \frac{1}{a}((ax + b) - b) = \frac{1}{a}(ax) = x
   $$

Thus, $$ (f \circ g)(x) = (g \circ f)(x) = x $$.

### 1.2.2: $$ f(x) = \frac{1}{x} $$, $$ g(x) = \frac{1}{x} $$

1. **Calculate $$ (f \circ g)(x) $$**:
   $$
   f(g(x)) = f\left(\frac{1}{x}\right) = \frac{1}{\frac{1}{x}} = x
   $$

2. **Calculate $$ (g \circ f)(x) $$**:
   $$
   g(f(x)) = g\left(\frac{1}{x}\right) = \frac{1}{\frac{1}{x}} = x
   $$

Thus, $$ (f \circ g)(x) = (g \circ f)(x) = x $$.

### 1.2.3: $$ f(x) = 2x $$, $$ g(x) = \frac{1}{2}x $$

1. **Calculate $$ (f \circ g)(x) $$**:
   $$
   f(g(x)) = f\left(\frac{1}{2}x\right) = 2\left(\frac{1}{2}x\right) = x
   $$

2. **Calculate $$ (g \circ f)(x) $$**:
   $$
   g(f(x)) = g(2x) = \frac{1}{2}(2x) = x
   $$

Thus, $$ (f \circ g)(x) = (g \circ f)(x) = x $$.

### 1.2.4: $$ f(x) = \frac{2x + 1}{x - 1} $$, $$ g(x) = \frac{x + 1}{x - 2} $$

1. **Calculate $$ (f \circ g)(x) $$**:
   $$
   f(g(x)) = f\left(\frac{x + 1}{x - 2}\right) = \frac{2\left(\frac{x + 1}{x - 2}\right) + 1}{\left(\frac{x + 1}{x - 2}\right) - 1}
   $$
   Simplifying the numerator:
   $$
   2\left(\frac{x + 1}{x - 2}\right) + 1 = \frac{2(x + 1) + (x - 2)}{x - 2} = \frac{2x + 2 + x - 2}{x - 2} = \frac{3x}{x - 2}
   $$
   Simplifying the denominator:
   $$
   \left(\frac{x + 1}{x - 2}\right) - 1 = \frac{x + 1 - (x - 2)}{x - 2} = \frac{3}{x - 2}
   $$
   Therefore,
   $$
   f(g(x)) = \frac{\frac{3x}{x - 2}}{\frac{3}{x - 2}} = x
   $$

2. **Calculate $$ (g \circ f)(x) $$**:
   $$
   g(f(x)) = g\left(\frac{2x + 1}{x - 1}\right) = \frac{\frac{2x + 1}{x - 1} + 1}{\frac{2x + 1}{x - 1} - 2}
   $$
   Simplifying the numerator:
   $$
   \frac{2x + 1}{x - 1} + 1 = \frac{2x + 1 + (x - 1)}{x - 1} = \frac{3x}{x - 1}
   $$
   Simplifying the denominator:
   $$
   \frac{2x + 1}{x - 1} - 2 = \frac{2x + 1 - 2(x - 1)}{x - 1} = \frac{2x + 1 - 2x + 2}{x - 1} = \frac{3}{x - 1}
   $$
   Therefore,
   $$
   g(f(x)) = \frac{\frac{3x}{x - 1}}{\frac{3}{x - 1}} = x
   $$

Thus, $$ (f \circ g)(x) = (g \circ f)(x) = x $$.

### Conclusion
For all cases, we have shown that $$ (f \circ g)(x) = (g \circ f)(x) = x $$.
For each pair of functions, the compositions $$ (f \circ g)(x) $$ and $$ (g \circ f)(x) $$ both equal $$ x $$.

1.2. For each of the following problems, show that

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