College Algebra: Piecewise Functions, Function Composition, and Related Concepts

Piecewise Functions and Their Properties

Definition and Construction of Piecewise Functions

A piecewise function is a function defined by different expressions over different intervals of its domain. Each 'piece' applies to a specific part of the domain, and the function may behave differently on each interval.

• Example: A function defined as a line for $x < 0$, a semicircle for $0 \leq x \leq 3$, and a constant for $x > 3$.

• General Form: $f(x) = \begin{cases} \text{expression 1}, & \text{if } x \in \text{interval 1} \\ \text{expression 2}, & \text{if } x \in \text{interval 2} \\ \vdots \end{cases}$

• Domain: The set of all $x$-values for which the function is defined.

• Range: The set of all possible output values ($y$-values) of the function.

Example: Suppose $f(x)$ is defined as follows:

• For $x < 0$: $f(x)$ is the line of slope $-2$ through $(0,1)$, so $f(x) = -2x + 1$.

• For $0 \leq x \leq 3$: $f(x)$ is the upper semicircle of radius $2$centered at$(1,0)$, so$f(x) = \sqrt{4 - (x-1)^2}$.

• For $x > 3$: $f(x) = 1$ (a constant function).

Domain: $(-\infty, \infty)$ (since each piece covers all real numbers). Range: $[0, 3]$ (since the semicircle's maximum is $2$, and the line and constant do not exceed $3$).

Function Composition and Piecewise Functions

Composing Functions

The composition of functions $(g \circ k)(x)$ means $g(k(x))$. When composing piecewise functions, the domain of the composition is determined by the domains of both functions and how their intervals overlap.

• Given: $q(x) = \begin{cases} x+2, & x < 0 \\ 2, & 0 \leq x \leq 2 \\ 4, & x > 2 \end{cases}$ and $k(x) = x-1$

• To find: $(q \circ k)(x) = q(k(x))$ as a piecewise function and its domain.

Steps:

1. Find the intervals for $x$ where $k(x)$ falls into each piece of $q$.

2. For $k(x) < 0 \implies x-1 < 0 \implies x < 1$: $q(k(x)) = (x-1) + 2 = x + 1$

3. For $0 \leq k(x) \leq 2 \implies 0 \leq x-1 \leq 2 \implies 1 \leq x \leq 3$: $q(k(x)) = 2$

4. For $k(x) > 2 \implies x-1 > 2 \implies x > 3$: $q(k(x)) = 4$

Result:

• $(q \circ k)(x) = \begin{cases} x+1, & x < 1 \\ 2, & 1 \leq x \leq 3 \\ 4, & x > 3 \end{cases}$

• Domain: $(-\infty, \infty)$

Rational Functions and Asymptotes

Finding Parameters for Intercepts and Asymptotes

A rational function is a function of the form $s(x) = \frac{Ax+B}{x-1} + C$. To find constants $A$, $B$, and $C$ given certain intercepts and asymptotes:

• x-intercept: Set $s(x) = 0$ and solve for $x$.

• y-intercept: Set $x = 0$ and solve for $s(0)$.

• Horizontal asymptote: Analyze the behavior as $x \to \infty$.

Example: Find $A$, $B$, $C$ so that $s(x) = \frac{Ax+B}{x-1} + C$ has:

• x-intercept at $x=2$

• y-intercept at $y=-1$

• Horizontal asymptote at $y=3$

Solution Outline:

1. Set $s(2) = 0$

2. Set $s(0) = -1$

3. As $x \to \infty$, $s(x) \to A + C = 3$

Solving Inequalities

Absolute Value and Rational Inequalities

• Absolute Value Inequality: $|2x-5| \leq 7$ Solution: $-7 \leq 2x-5 \leq 7$ Solve for $x$ to find the solution interval.

• Rational Inequality: $\left|\frac{x+1}{x-2}\right| > 2$ (excluding points not in the domain, i.e., $x \neq 2$)

Key Steps:

1. For absolute value: Split into two inequalities.

2. For rational: Consider where the expression is defined and solve both $\frac{x+1}{x-2} > 2$ and $\frac{x+1}{x-2} < -2$.

Range of Quadratic Functions

Finding the Range

To find the exact range of a quadratic function $f(x) = -2x^2 + 8x - 5$:

• Find the vertex: $x_v = -\frac{b}{2a}$

• Compute $f(x_v)$ for the maximum or minimum value (since $a < 0$, the parabola opens downward and the vertex is a maximum).

• Range: $(-\infty, f(x_v)]$

Function Composition and Domains

Domain of Composed Functions

• Given $u(x) = \sqrt{2-x}$ and $v(x) = \frac{1}{x-4}$

• (a) $(u \circ v)(x) = u(v(x))$: Substitute $v(x)$ into $u$ and find the domain where $2 - v(x) \geq 0$ and $v(x)$ is defined.

• (b) $(v \circ u)(x) = v(u(x))$: Substitute $u(x)$ into $v$ and find the domain where $u(x)$ is defined and $u(x) \neq 4$.

• (c) Domain Differences: The domain of a composition depends on both the inner and outer functions. Restrictions may arise from either function.

Defining Piecewise Linear Functions

Constructing a Piecewise Linear Function with Given Values

To define a function $T$ so that $T(-2) = 3$, $T(0) = 0$, $T(4) = -2$, and $T$ is linear on each of the intervals $(-\infty, -2]$, $[-2, 0]$, $[0, 4]$, $[4, \infty)$ with constant values outside $[-2, 4]$:

• For $x \leq -2$: $T(x) = 3$

• For $-2 < x \leq 0$: $T(x)$ is linear from $( -2, 3 )$ to $( 0, 0 )$

• For $0 < x \leq 4$: $T(x)$ is linear from $( 0, 0 )$ to $( 4, -2 )$

• For $x > 4$: $T(x) = -2$

Explicit Formula:

• For $-2 < x \leq 0$: $T(x) = \frac{-3}{2}(x + 2) + 3$

• For $0 < x \leq 4$: $T(x) = -\frac{1}{2}x$

Sketch: The graph consists of two line segments connecting the given points, with constant values outside $[-2, 4]$.