Section 1.3: New Functions from Old Functions

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MATH161

Skills in This Section

What is This Section About?

Imagine you’re an artist with a basic shape—say, a circle. By shifting, stretching, reflecting, and combining it with other shapes, you can create infinitely many new designs. This is exactly what we do with functions!

In this section, you’ll learn the powerful techniques for creating new functions from existing ones. Instead of memorizing hundreds of different graphs, you’ll master a handful of transformations that let you quickly sketch complex functions. You’ll also learn to combine functions through arithmetic operations and composition—skills that are absolutely essential for calculus.

Why does this matter? Because calculus is all about understanding how functions change. When you see $f(x-3)$ or $2f(x)$ or $f(g(x))$, you need to instantly visualize what’s happening. These transformation skills are your shortcut to understanding complex functions without tedious calculations or plotting dozens of points.

The Big Picture: What You’ll Learn

By the end of this section, you’ll be able to:

1. Apply vertical and horizontal shifts to translate function graphs
2. Use stretches and compressions to scale functions vertically and horizontally
3. Perform reflections across the x-axis and y-axis
4. Transform functions using absolute values to manipulate negative portions
5. Combine functions through addition, subtraction, multiplication, and division
6. Compose functions to create complex relationships from simple building blocks
7. Determine domains of combined and composed functions
8. Decompose complex functions into simpler components

Think of this as your “function manipulation toolkit”—once you master these techniques, you’ll be able to understand and sketch a vast array of functions quickly and confidently.

Core Concepts

Vertical Shifts: Moving Up and Down

Rule: $y = f(x) + c$

• If $c > 0$: shift the graph upward by $c$ units
• If $c < 0$: shift the graph downward by $\vert c\vert $ units

In plain English: Adding a constant outside the function lifts the entire graph (or drops it if negative). Every point moves the same vertical distance.

Example:

• Start with $y = x^2$ (a parabola with vertex at the origin)
• $y = x^2 + 3$ shifts it up 3 units (vertex now at $(0, 3)$)
• $y = x^2 - 2$ shifts it down 2 units (vertex now at $(0, -2)$)

Key Insight: Vertical shifts change the range but not the domain.

Horizontal Shifts: Moving Left and Right

Rule: $y = f(x - c)$

• If $c > 0$: shift the graph right by $c$ units (OPPOSITE of what you expect!)
• If $c < 0$: shift the graph left by $\vert c\vert $ units

The “opposite sign” rule: This is the most commonly confused transformation! Remember:

• $f(x - 3)$ means “3 to the RIGHT”
• $f(x + 3)$ means “3 to the LEFT”

Why? Think about when the function equals zero. For $f(x-3)$ to equal $f(0)$, we need $x-3=0$, so $x=3$. The “center” has moved right to $x=3$.

Example:

• Start with $y = \sqrt{x}$ (starts at origin, goes right)
• $y = \sqrt{x-4}$ shifts it right 4 units (now starts at $(4, 0)$)
• $y = \sqrt{x+2}$ shifts it left 2 units (now starts at $(-2, 0)$)

Key Insight: Horizontal shifts change the domain but not the range.

Vertical Stretches and Compressions

Rule: $y = cf(x)$ where $c > 0$

• If $c > 1$: stretch vertically by factor $c$ (graph gets taller)
• If $0 < c < 1$: compress vertically by factor $c$ (graph gets shorter)

What’s happening: Every $y$-value gets multiplied by $c$. If $c = 2$, every point that was at height 5 is now at height 10.

Example:

• Start with $y = \sin x$ (oscillates between -1 and 1)
• $y = 3\sin x$ stretches it (now oscillates between -3 and 3)
• $y = \frac{1}{2}\sin x$ compresses it (now oscillates between -0.5 and 0.5)

Horizontal Stretches and Compressions

Rule: $y = f(cx)$ where $c > 0$

• If $c > 1$: compress horizontally by factor $c$ (graph gets narrower)
• If $0 < c < 1$: stretch horizontally by factor $1/c$ (graph gets wider)

CAUTION: This is opposite of vertical stretches! Larger $c$ means horizontal compression.

Why? For $f(2x)$ to reach a value that $f(x)$ reaches at $x=10$, we only need $x=5$ (since $2 \cdot 5 = 10$). Things happen “faster,” so the graph compresses.

Example:

• Start with $y = \sin x$ (period is $2\pi$)
• $y = \sin(2x)$ compresses it (period becomes $\pi$)
• $y = \sin\left(\frac{x}{2}\right)$ stretches it (period becomes $4\pi$)

Reflections: Flipping the Graph

Two types:

Reflection about the x-axis: $y = -f(x)$

• Flip the graph upside down
• Positive $y$-values become negative, negative become positive
• Example: $y = -x^2$ is an upside-down parabola

Reflection about the y-axis: $y = f(-x)$

• Flip the graph left-to-right
• What was on the right is now on the left
• Example: $y = \sqrt{-x}$ is the square root function reflected (opens to the left)

Thought Process:

• Negative outside: flip vertically ($-f(x)$)
• Negative inside: flip horizontally ($f(-x)$)

Combining reflections:

• $y = -f(-x)$ reflects about both axes (equivalent to 180° rotation about origin)

Absolute Value Transformations

Rule: $y = \vert f(x)\vert $

Effect: Take any portion of the graph that’s below the x-axis and reflect it upward.

Step-by-step:

1. Draw the original function $y = f(x)$
2. Keep everything above the x-axis as is
3. Flip everything below the x-axis up (make it positive)

Example: $y = \vert x - 2\vert $

• Original $y = x - 2$ is a line with negative values when $x < 2$
• $y = \vert x - 2\vert $ creates a V-shape with vertex at $(2, 0)$

Key Insight: Absolute value functions always have range $[0, \infty)$ since they can’t produce negative outputs.

Combining Functions Arithmetically

Given two functions $f$ and $g$, we can create new functions:

Sum: $(f + g)(x) = f(x) + g(x)$

Difference: $(f - g)(x) = f(x) - g(x)$

Product: $(f \cdot g)(x) = f(x) \cdot g(x)$

Quotient: $\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}$, where $g(x) \neq 0$

Domain rules:

• For sum, difference, and product: domain is the intersection of the individual domains
• For quotient: domain is the intersection minus any points where $g(x) = 0$

Example:

• $f(x) = \sqrt{x}$ has domain $[0, \infty)$
• $g(x) = \sqrt{4-x}$ has domain $(-\infty, 4]$
• $(f + g)(x) = \sqrt{x} + \sqrt{4-x}$ has domain $[0, 4]$ (the intersection)

Composition of Functions: Functions of Functions

Definition: $(f \circ g)(x) = f(g(x))$

In plain English: Feed $x$ into $g$ first, then feed the result into $f$. Think of it as a two-stage process.

ORDER MATTERS! Usually $(f \circ g)(x) \neq (g \circ f)(x)$

Example:

• Let $f(x) = x^2$ and $g(x) = x + 1$
• $(f \circ g)(x) = f(g(x)) = f(x+1) = (x+1)^2 = x^2 + 2x + 1$
• $(g \circ f)(x) = g(f(x)) = g(x^2) = x^2 + 1$
• These are different functions!

Domain of composition: The domain of $f \circ g$ consists of all $x$ in the domain of $g$ such that $g(x)$ is in the domain of $f$.

Thought process for finding domain:

1. Find where $g(x)$ is defined
2. Find where $f$ is defined
3. Keep only the $x$-values where $g(x)$ stays within $f$’s domain

Decomposing Functions

Sometimes you need to work backward: given a complex function, break it into simpler pieces.

Strategy: Look for the “inside” function (innermost operation) first.

Example: Decompose $h(x) = \sqrt{x^2 + 1}$

Solution:

• Inner function: $g(x) = x^2 + 1$ (what’s inside the square root)
• Outer function: $f(x) = \sqrt{x}$ (the square root operation)
• Check: $(f \circ g)(x) = f(g(x)) = \sqrt{x^2 + 1}$ ✓

More complex example: Decompose $h(x) = [\cos(x + 9)]^2$

Solution:

• Innermost: $u(x) = x + 9$ (the shift)
• Middle: $v(x) = \cos x$ (the cosine)
• Outer: $w(x) = x^2$ (the squaring)
• Check: $(w \circ v \circ u)(x) = [\cos(x+9)]^2$ ✓

Worked Examples

Example 1: Multiple Transformations

Problem: Start with $y = \sqrt{x}$ and describe the transformations needed to obtain $y = -2\sqrt{x-3} + 1$.

Thought Process: I need to work through this systematically, identifying each transformation in order.

Solution:

Starting function: $y = \sqrt{x}$

Step 1: Horizontal shift right 3 units \(y = \sqrt{x-3}\)

Step 2: Vertical stretch by factor 2 \(y = 2\sqrt{x-3}\)

Step 3: Reflection about x-axis \(y = -2\sqrt{x-3}\)

Step 4: Vertical shift up 1 unit \(y = -2\sqrt{x-3} + 1\)

Summary of transformations:

1. Shift right 3 (domain changes from $[0,\infty)$ to $[3,\infty)$)
2. Stretch vertically by 2 (makes it rise faster)
3. Reflect over x-axis (flips it upside down)
4. Shift up 1 (raises the entire graph)

Key Takeaway: Work systematically: inside operations (horizontal) first, then outside operations (vertical). The order matters!

Example 2: Graphing with Transformations

Problem: Sketch $y = \vert x^2 - 4\vert $ without making a table of values.

Thought Process: I’ll first sketch $y = x^2 - 4$, then apply the absolute value transformation.

Solution:

Step 1: Sketch $y = x^2 - 4$

• This is a parabola shifted down 4 units
• Vertex at $(0, -4)$
• Crosses x-axis where $x^2 - 4 = 0$, so $x = \pm 2$

Step 2: Apply absolute value $y = \vert x^2 - 4\vert $

• The portion where $x^2 - 4 \geq 0$ (i.e., $x \leq -2$ or $x \geq 2$) stays the same
• The portion where $x^2 - 4 < 0$ (i.e., $-2 < x < 2$) gets reflected upward

Result:

• A W-shaped graph
• The bottom of the parabola (from $x=-2$ to $x=2$) that was below the x-axis is now reflected upward
• Minimum value is 0 (at $x = \pm 2$)
• Vertex moves from $(0, -4)$ to $(0, 4)$

Key Takeaway: Absolute value transformations create distinctive “sharp corners” where the original function crosses the x-axis.

Example 3: Composition of Functions

Problem: If $f(x) = x^2$ and $g(x) = x - 3$, find: (a) $(f \circ g)(x)$ and $(g \circ f)(x)$ (b) Show that these are different functions

Thought Process: I’ll carefully apply the definition of composition, making sure to substitute correctly.

Solution:

(a) Find $(f \circ g)(x) = f(g(x))$

Step 1: Start with $g(x) = x - 3$

Step 2: Substitute into $f$: \(f(g(x)) = f(x-3) = (x-3)^2 = x^2 - 6x + 9\)

(b) Find $(g \circ f)(x) = g(f(x))$

Step 1: Start with $f(x) = x^2$

Step 2: Substitute into $g$: \(g(f(x)) = g(x^2) = x^2 - 3\)

(c) Compare:

• $(f \circ g)(x) = x^2 - 6x + 9$
• $(g \circ f)(x) = x^2 - 3$

These are clearly different! For example, when $x = 0$:

• $(f \circ g)(0) = 9$
• $(g \circ f)(0) = -3$

Key Takeaway: Composition is NOT commutative. The order in which you apply functions matters tremendously!

Example 4: Finding Domain of Composition

Problem: Find the domain of $(f \circ g)(x)$ where $f(x) = \sqrt{x}$ and $g(x) = 4 - x^2$.

Thought Process: I need $g(x)$ to be defined AND its output must be in the domain of $f$.

Solution:

Step 1: Find $(f \circ g)(x)$ \((f \circ g)(x) = f(g(x)) = f(4-x^2) = \sqrt{4-x^2}\)

Step 2: Determine restrictions

• $g(x) = 4 - x^2$ is defined for all real $x$ ✓
• BUT $f(x) = \sqrt{x}$ requires $x \geq 0$
• So we need $g(x) \geq 0$

Step 3: Solve $4 - x^2 \geq 0$ \(4 \geq x^2\) \(x^2 \leq 4\) \(-2 \leq x \leq 2\)

Domain: $[-2, 2]$

Key Takeaway: For composition, always check that the output of the inner function falls within the domain of the outer function. Don’t just assume everything works!

Practice Problems

Try these problems to solidify your understanding:

1. Vertical Shifts: Write the equation for $y = x^2$ shifted up 5 units.

2. Horizontal Shifts: Write the equation for $y = \sqrt{x}$ shifted left 2 units.

3. Reflections: How do you transform $y = x^3$ to reflect it about the y-axis?

4. Stretches: Describe the transformation from $y = \cos x$ to $y = 4\cos x$.

5. Multiple Transformations: Describe all transformations from $y = x^2$ to $y = -\frac{1}{2}(x+3)^2 - 1$.

6. Composition: If $f(x) = \frac{1}{x}$ and $g(x) = x^2 + 1$, find $(f \circ g)(x)$ and its domain.

7. Decomposition: Express $h(x) = \sqrt{x^2 + 5}$ as a composition $f \circ g$.

8. Absolute Value: Sketch $y = \vert x - 1\vert $ and identify its domain and range.

Hints:

• Problem 1: Add the constant outside
• Problem 2: Remember the “opposite sign” rule!
• Problem 3: Replace $x$ with $-x$
• Problem 4: This is a vertical stretch
• Problem 5: Work from inside out: horizontal shift, vertical stretch, reflection, vertical shift
• Problem 6: Make sure $g(x)$ never equals zero
• Problem 7: What’s the innermost operation?
• Problem 8: Where does the function change direction?

Key Reminders

As you work with function transformations, remember:

✓ Inside vs. Outside – Changes inside $f(\cdot)$ affect horizontal behavior; changes outside affect vertical

✓ Opposite sign rule – $f(x-c)$ shifts RIGHT (opposite of intuition!)

✓ Order matters for composition – $(f \circ g) \neq (g \circ f)$ in general

✓ Reflections: minus inside vs. outside – $-f(x)$ flips vertically, $f(-x)$ flips horizontally

✓ Domain of compositions – Check that inner function’s output is in outer function’s domain

✓ Absolute value creates corners – Look for where the original function crosses the x-axis

Why This Matters

Function transformations are everywhere in the real world:

• Signal processing: Audio engineers shift and scale waveforms
• Economics: Cost functions are often transformations of production functions
• Physics: Wave functions involve shifts, stretches, and compositions
• Computer graphics: Every animated character involves transforming basic shapes

More importantly, composition is the foundation of the Chain Rule in calculus—one of the most powerful tools you’ll learn. Understanding $f(g(x))$ now will make derivatives of complex functions much easier later!