12.4 Function

1. Find \(f \circ g\). Find the domain and range:

   1. \(f(x) = x^2 + 2x, \quad g(x) = 2x + 1\)

      Solution:
      ​\(\text{Dom}(f \circ g) = \{x \in \text{Dom}(g) \mid g(x) \in \text{Dom}(f)\} = \mathbb{R}\)

      \(f(g(x)) = f(2x + 1) = (2x + 1)^2 + 2(2x + 1)\)​\(= 4x^2 + 4x + 1 + 4x + 2 = 4x^2 + 8x + 3\) 2. \(f(x)=\begin{cases}x+1 & \text{if }x\leq0\\ 2x & \text{if }x>0\end{cases},\quad g(x)=\begin{cases}x & \text{if }x\leq-1\\ -x & \text{if }x>-1\end{cases}\)​

      \(f(g(x))=\begin{cases}f(x) & \text{if }x\leq-1\\ f(-x) & \text{if }x>-1\end{cases}\) \(f(g(x))=\begin{cases}2x+1 & \text{if }x\leq-1\\ -2x & \text{if }x\in(-1,0)\\ -x+1 & \text{if }x\in[0,\infty)\end{cases}\)​\(\text{Dom}(f \circ g) = \mathbb{R}\)

2. Prove or give a counterexample:

   1. If \(f\) is a linear function with positive slope \(\implies f\) is increasing in \(\mathbb{R}\).

      T 2. If \(\text{Dom}(f) = \mathbb{R}\) and \(f\) is increasing on the intervals \([-2, -1]\) and \([1, 2]\), \(\implies f\) is increasing on \([-2, 2]\)

      F 3. If \(f\) is increasing on \([1, 2]\) and also increasing on \((2, 3]\), then \(f\) is increasing in \([1, 3]\)

      F 4. If \(f\) is increasing on \((-\infty, 0)\) and increasing on \([0, \infty)\), \(\implies f\) is increasing on \(\mathbb{R}\)

      F

3. Determine if \(f\) is surjective/injective/bijective:

   1. The identity function from \(A\) to \(A\):

      \(\text{inj and surj, so bijective.}\) 2. The inclusion function from \(A\) to \(X\), \(A \subseteq X\):
      ​\(\text{inj, not surjective because } A \neq X.\) 3. The canonical map from \(\mathbb{Z} \to \mathbb{Z}_5\):
      ​\(\text{Not injective because } f(0) = 0 = 5 = f(5).\) 4. The greatest integer function \(\text{int}: \mathbb{R} \to \mathbb{Z}\):
      ​\(\text{It is not injective because } \text{int}(2.1) = 2 = \text{int}(2).\)​\(\text{It is surjective because: let } n \in \mathbb{Z}, \text{int}(n) = n.\)

4. Prove that for \(f: A \to B\), \(g: B \to C\), if \(g \circ f: A \to C\) is injective \(\implies f: A \to B\) is injective

   Proof:

   Suppose \(x_1 \neq x_2\) but \(f(x_1) = f(x_2)\) (not injective).

   \(g(f(x_1)) = g(f(x_2)) \implies (g \circ f)(x_1) = (g \circ f)(x_2)\)

   This is impossible. Hence, \(f\) must be injective.

5. Prove that \(f(x)=\begin{cases}\frac{x-2}{x+4} & \text{if }x\neq-4\\ 1 & \text{if }x=-4\end{cases}\text{ is onto }\mathbb{R}\text{ and }1-1.\)​

   Let's see that \(f\) is injective:

   Let \(x_1\) and \(x_2\) be real numbers.

   Case 1: \(x_1 \neq -4\) and \(x_2 \neq -4\)

   \(\frac{x_1 - 2}{x_1 + 4} = \frac{x_2 - 2}{x_2 + 4} \implies (x_1 - 2)(x_2 + 4) = (x_2 - 2)(x_1 + 4)\)

   \(\implies x_1 x_2 + 4x_1 - 2x_2 - 8 = x_2 x_1 + 4x_2 - 2x_1 - 8\)

   \(\implies 6x_1 = 6x_2 \implies x_1 = x_2\)

   Case 2: \(x_1 = -4\) and \(x_2 \neq -4\)

   \(f(x_1) = 1\), \(f(x_2) = \frac{x_2 - 2}{x_2 + 4}\)

   \(\frac{x_2 - 2}{x_2 + 4} = 1 \implies x_2 - 2 = x_2 + 4 \implies -2 = 4\) (a contradiction)

   So, if \(x_1 = -4\) and \(x_2 \neq -4\), \(f(x_1) \neq f(x_2)\).

   In conclusion, \(f\) is injective in \(\mathbb{R}\).

   To prove \(f\) is surjective:

   Let \(y \in \mathbb{R}\).

   If \(y = 1 \implies f(-4) = 1\).

   If \(y \neq 1\), then \(f\left(\frac{-2\left(2y+1\right)}{y-1}\right)=\ldots=y\)

   Therefore, \(f\) is surjective.

   In conclusion, \(f\) is bijective.

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