1. **Problem Statement:** Define the following terms with examples: (a) Function, (b) Even function, (c) Odd function, (d) Domain, (e) Range. 2. **Definitions and Examples:** 1. (a) **Function:** A function $f$ from a set $A$ to a set $B$ assigns each element in $A$ exactly one element in $B$. Example: $f(x) = 2x + 3$. 2. (b) **Even function:** A function $f$ is even if $f(-x) = f(x)$ for all $x$ in its domain. Example: $f(x) = x^2$. 3. (c) **Odd function:** A function $f$ is odd if $f(-x) = -f(x)$ for all $x$ in its domain. Example: $f(x) = x^3$. 4. (d) **Domain:** The set of all possible input values $x$ for which the function is defined. 5. (e) **Range:** The set of all possible output values $f(x)$ of the function. --- **Next, analyze the function:** (a) $f(x) = |x| + |x - 1|$ - **Domain:** All real numbers $x$ because absolute value is defined everywhere. - **Range:** Since both terms are non-negative, minimum occurs at some $x$. Check critical points: For $x < 0$: $f(x) = -x + 1 - x = 1 - 2x$ For $0 \\leq x < 1$: $f(x) = x + 1 - x = 1$ For $x \\geq 1$: $f(x) = x + x - 1 = 2x - 1$ Minimum value is $1$ at $x$ in $[0,1]$. Range is $[1, \infty)$. (b) $f(x) = |x + 1| + |x - 2|$ - **Domain:** All real numbers. - **Range:** Minimum occurs between $-1$ and $2$. For $x < -1$: $f(x) = -x - 1 + 2 - x = 1 - 2x$ For $-1 \\leq x < 2$: $f(x) = x + 1 + 2 - x = 3$ For $x \\geq 2$: $f(x) = x + 1 + x - 2 = 2x - 1$ Minimum value is $3$ at $x$ in $[-1,2]$. Range is $[3, \infty)$. (c) Piecewise function: $$ f(x) = \begin{cases} x^2 & x < 0 \\ x & 0 \leq x \leq 1 \\ \frac{1}{x} & x > 1 \end{cases} $$ - **Domain:** All real numbers except $x=0$ is included since $x=0$ is in the middle piece. - **Range:** For $x<0$, $x^2 \geq 0$. For $0 \leq x \leq 1$, $f(x) = x$ ranges from $0$ to $1$. For $x > 1$, $f(x) = \frac{1}{x}$ ranges from $1$ down to $0$ (not including 0). Range is $[0, \infty)$. (d) Piecewise function: $$ f(x) = \begin{cases} x^2 + 1 & x < 0 \\ x & 0 \leq x \leq 1 \\ \frac{1}{x} & x > 1 \end{cases} $$ - **Domain:** All real numbers except $x=0$ is included. - **Range:** For $x<0$, $x^2 + 1 \geq 1$. For $0 \leq x \leq 1$, $f(x) = x$ ranges from $0$ to $1$. For $x > 1$, $f(x) = \frac{1}{x}$ ranges from $1$ down to $0$ (not including 0). Range is $(0, \infty)$. (e) Piecewise function: $$ f(x) = \begin{cases} 0 & |x| > 1 \\ 1 + x & -1 \leq x \leq 0 \\ 1 - x & 0 < x \leq 1 \end{cases} $$ - **Domain:** All real numbers. - **Range:** For $|x| > 1$, $f(x) = 0$. For $-1 \leq x \leq 0$, $f(x) = 1 + x$ ranges from $0$ to $1$. For $0 < x \leq 1$, $f(x) = 1 - x$ ranges from $1$ down to $0$. Range is $[0,1]$. --- **Summary:** - (a) Domain: $(-\infty, \infty)$, Range: $[1, \infty)$ - (b) Domain: $(-\infty, \infty)$, Range: $[3, \infty)$ - (c) Domain: $(-\infty, \infty)$, Range: $[0, \infty)$ - (d) Domain: $(-\infty, \infty)$, Range: $(0, \infty)$ - (e) Domain: $(-\infty, \infty)$, Range: $[0,1]$