MA161 - Lesson 1: Review of Functions

Definition: A function is a rule that assigns exactly one output for each input.

Notation: \(y = f(x)\), where \(x\) is the input and \(y\) is the output.

Example: Function Evaluation

Given \(f(x) = x^3 - 5x + 1\), find the following outputs:

Solution:

Input = 1: \(f(1) = 1^3 - 5(1) + 1 = 1 - 5 + 1 = -3\)
Input = 0: \(f(0) = 0^3 - 5(0) + 1 = 0 - 0 + 1 = 1\)
Input = \(a\): \(f(a) = a^3 - 5a + 1\)

Vertical Line Test

Graph Analysis: The handwritten notes show two coordinate plane graphs. Graph 1 shows a curve that passes the vertical line test (each vertical line intersects the curve at most once), while Graph 2 shows a curve that fails the test (some vertical lines intersect multiple times).

Vertical Line Test: A graph represents a function if and only if every vertical line intersects the graph at most once.

On a graph: horizontal axis represents inputs, vertical axis represents outputs.

Domain and Range

Domain: All acceptable inputs for a function

Range: All possible outputs from a function

Example: Finding Domain and Range

For \(f(x) = \sqrt{x + 5}\):

Finding Domain:

For the square root to be defined, we need: \(x + 5 \geq 0\)

Therefore: \(x \geq -5\)

Domain: \([-5, \infty)\)

Finding Range:

Testing values:

Input = 4 → Output = \(\sqrt{9} = 3\)
Input = -1 → Output = \(\sqrt{4} = 2\)
Input = -4 → Output = \(\sqrt{1} = 1\)
Input = -5 → Output = \(\sqrt{0} = 0\)
Input = -6 → Not acceptable (would give \(\sqrt{-1}\))

Range: \([0, \infty)\)

Example: More Complex Domain

Find the domain of \(f(x) = \sqrt{x-1} + \frac{4}{\sqrt{2-x}}\)

Solution:

For this function to be defined, we need both parts to be acceptable:

For \(\sqrt{x-1}\): \(x - 1 \geq 0\), so \(x \geq 1\)
For \(\frac{4}{\sqrt{2-x}}\): \(2 - x > 0\), so \(x < 2\)

Combining both conditions: \(1 \leq x < 2\)

Domain: \([1, 2)\)

Number Line Visualization: The handwritten notes include a number line showing the interval [1, 2) with a closed circle at x = 1 and an open circle at x = 2, illustrating the domain graphically.

Composition of Functions

Function Composition: Given functions \(f(x)\) and \(g(x)\), the composition is written as:

\((f \circ g)(x) = f(g(x))\)
\((g \circ f)(x) = g(f(x))\)

Important: \(f(g(x))\) need not equal \(g(f(x))\)

Example: Composition of Functions

Given \(f(x) = \frac{1}{x}\) and \(g(x) = \sqrt{2-x}\):

Solution:

\(f(g(x)) = f(\sqrt{2-x}) = \frac{1}{\sqrt{2-x}}\)

\(g(f(x)) = g\left(\frac{1}{x}\right) = \sqrt{2-\frac{1}{x}}\)

Note: These are different functions, showing that composition is not commutative.

Symmetry

Even Functions: \(f(-x) = f(x)\) for all \(x\) in the domain

Example: \(f(x) = x^2\) since \(f(-x) = (-x)^2 = x^2 = f(x)\)

Odd Functions: \(f(-x) = -f(x)\) for all \(x\) in the domain

Example: \(f(x) = x^3\) since \(f(-x) = (-x)^3 = -x^3 = -f(x)\)

Transformations

Graph Transformations: The handwritten notes show multiple curves illustrating various transformations of a base function, with different colors representing different transformations including shifts, stretches, and reflections.

Vertical Transformations

Vertical Shift:

\(y = f(x) + k\): shifts up \(k\) units if \(k > 0\), down if \(k < 0\)

Vertical Stretch/Compression:

\(y = kf(x)\): stretches vertically if \(|k| > 1\), compresses if \(0 < |k| < 1\)

Horizontal Transformations

Horizontal Shift:

\(y = f(x - k)\): shifts right \(k\) units if \(k > 0\), left if \(k < 0\)

Horizontal Stretch/Compression:

\(y = f(kx)\): compresses horizontally if \(|k| > 1\), stretches if \(0 < |k| < 1\)

Reflections

Reflections:

\(y = f(-x)\): reflects about the y-axis (horizontal reflection)
\(y = -f(x)\): reflects about the x-axis (vertical reflection)

Reflection Diagrams: The handwritten notes include coordinate plane graphs showing how functions are reflected across both axes, with the original function and reflected versions clearly marked.

Example: Combined Transformations

Sketch the graph of \(F(x) = 6(x+5)^2 + 2\) using transformations of \(y = x^2\).

Solution:

Starting with \(y = x^2\), apply transformations in order:

Horizontal shift left 5 units: \(y = (x+5)^2\)
Vertical stretch by factor 6: \(y = 6(x+5)^2\)
Vertical shift up 2 units: \(y = 6(x+5)^2 + 2\)

The vertex moves from (0,0) to (-5,2).

Transformation Example Graph: The handwritten notes show a parabola that has been shifted and stretched, with the vertex clearly marked at (-5, 2) and the shape wider than the original parabola y = x².