Solved Exercises on Function Notation

Master function notation: f(x), evaluating functions, domain, range, and function operations through these 5 detailed exercises.

Solution: Exercises 1 to 3
1 Basic Function Evaluation
Exercise 1
Given f(x) = 2x + 3, evaluate f(5), f(-2), and f(a + 1).
Definition:

Function notation: f(x) represents the output when x is the input

Evaluation: Replace every instance of x with the given value

Method for function evaluation:
  1. Identify the function rule f(x)
  2. Replace every occurrence of x with the given input value
  3. Simplify the resulting expression
  4. Calculate the final result
Function
f(x) = 2x + 3
Evaluation
f(5) = 2(5) + 3
Step 1: Evaluate f(5)

f(x) = 2x + 3

f(5) = 2(5) + 3 = 10 + 3 = 13

Step 2: Evaluate f(-2)

f(x) = 2x + 3

f(-2) = 2(-2) + 3 = -4 + 3 = -1

Step 3: Evaluate f(a + 1)

f(x) = 2x + 3

f(a + 1) = 2(a + 1) + 3 = 2a + 2 + 3 = 2a + 5

Step 4: Verify calculations

f(5) = 13: 2(5) + 3 = 10 + 3 = 13 ✓

f(-2) = -1: 2(-2) + 3 = -4 + 3 = -1 ✓

f(a + 1) = 2a + 5: 2(a + 1) + 3 = 2a + 2 + 3 = 2a + 5 ✓

f(5) = 13
f(-2) = -1
f(a + 1) = 2a + 5
Final answer:

f(5) = 13

f(-2) = -1

f(a + 1) = 2a + 5

Applied rules:

• Function evaluation: Substitute input value for x everywhere in the function

• Distribution: Apply distributive property when substituting expressions

• Simplification: Combine like terms after substitution

2 Function Operations
Exercise 2
Given f(x) = x² - 2x + 1 and g(x) = 3x - 4, find (f + g)(x), (f - g)(x), and (f · g)(x).
Definition:

Function operations: (f + g)(x) = f(x) + g(x), (f - g)(x) = f(x) - g(x), (f · g)(x) = f(x) · g(x)

Combining functions: Perform operation on the output values

Given
f(x) = x² - 2x + 1, g(x) = 3x - 4
Addition
(f + g)(x) = x² + x - 3
Subtraction
(f - g)(x) = x² - 5x + 5
Step 1: Find (f + g)(x)

(f + g)(x) = f(x) + g(x)

= (x² - 2x + 1) + (3x - 4)

= x² - 2x + 1 + 3x - 4

= x² + x - 3

Step 2: Find (f - g)(x)

(f - g)(x) = f(x) - g(x)

= (x² - 2x + 1) - (3x - 4)

= x² - 2x + 1 - 3x + 4

= x² - 5x + 5

Step 3: Find (f · g)(x)

(f · g)(x) = f(x) · g(x)

= (x² - 2x + 1)(3x - 4)

= x²(3x - 4) - 2x(3x - 4) + 1(3x - 4)

= 3x³ - 4x² - 6x² + 8x + 3x - 4

= 3x³ - 10x² + 11x - 4

Step 4: Verify with specific values

Let x = 1: f(1) = 0, g(1) = -1

(f + g)(1) = 0 + (-1) = -1

Using formula: 1² + 1 - 3 = -1 ✓

(f + g)(x) = x² + x - 3
(f - g)(x) = x² - 5x + 5
(f · g)(x) = 3x³ - 10x² + 11x - 4
Final answer:

(f + g)(x) = x² + x - 3

(f - g)(x) = x² - 5x + 5

(f · g)(x) = 3x³ - 10x² + 11x - 4

Applied rules:

• Function addition: Add corresponding outputs: (f + g)(x) = f(x) + g(x)

• Function subtraction: Subtract outputs: (f - g)(x) = f(x) - g(x)

• Function multiplication: Multiply outputs: (f · g)(x) = f(x) · g(x)

3 Composite Functions
Exercise 3
Given f(x) = 2x + 1 and g(x) = x² - 3, find (f ∘ g)(x) and (g ∘ f)(x).
Definition:

Composite function: (f ∘ g)(x) = f(g(x)), read as "f of g of x"

Order matters: Generally (f ∘ g)(x) ≠ (g ∘ f)(x)

Given
f(x) = 2x + 1, g(x) = x² - 3
Composition 1
(f ∘ g)(x) = 2x² - 5
Composition 2
(g ∘ f)(x) = 4x² + 4x - 2
Step 1: Find (f ∘ g)(x)

(f ∘ g)(x) = f(g(x))

First, find g(x): g(x) = x² - 3

Then, substitute into f: f(g(x)) = f(x² - 3)

f(x² - 3) = 2(x² - 3) + 1 = 2x² - 6 + 1 = 2x² - 5

Step 2: Find (g ∘ f)(x)

(g ∘ f)(x) = g(f(x))

First, find f(x): f(x) = 2x + 1

Then, substitute into g: g(f(x)) = g(2x + 1)

g(2x + 1) = (2x + 1)² - 3 = 4x² + 4x + 1 - 3 = 4x² + 4x - 2

Step 3: Compare results

(f ∘ g)(x) = 2x² - 5

(g ∘ f)(x) = 4x² + 4x - 2

These are different functions, confirming order matters in composition!

Step 4: Verify with specific value

Let x = 2:

g(2) = 4 - 3 = 1, f(1) = 2 + 1 = 3

So (f ∘ g)(2) = 3

Using formula: 2(4) - 5 = 8 - 5 = 3 ✓

(f ∘ g)(x) = 2x² - 5
(g ∘ f)(x) = 4x² + 4x - 2
Final answer:

(f ∘ g)(x) = 2x² - 5

(g ∘ f)(x) = 4x² + 4x - 2

Applied rules:

• Composition definition: (f ∘ g)(x) = f(g(x))

• Order matters: Function composition is generally not commutative

• Substitution: Replace inner function's output as input to outer function

Function Notation Summary: Definitions, Rules, and Methods
f(x) = y
Function Notation
Basic Notation
f(x)
Read as "f of x"
Operations
(f + g)(x)
Addition of functions
Composition
(f ∘ g)(x)
Function of a function
Key definitions:

Function: A relation where each input has exactly one output

Domain: Set of all possible input values (x-values)

Range: Set of all possible output values (y-values)

Function notation: f(x) represents the output when x is the input

Function evaluation methodology:
  1. Identify the function: Determine f(x) rule
  2. Substitute: Replace x with the given value
  3. Simplify: Perform arithmetic operations
  4. Calculate: Obtain final result
Tip 1: f(x) is just another way to write y, emphasizing the input-output relationship.
Tip 2: Always substitute the entire input expression for x.
Tip 3: For compositions, work from inside out: f(g(x)) means evaluate g first, then f.
Tip 4: Check your work by substituting a simple value to verify.
Common errors: Forgetting to substitute all x's, incorrect distribution, mixing up order in compositions.
Key translations: f(x) = y, f(a) = output when input is a, f(x+h) = f evaluated at (x+h).
Essential formulas and rules:

• Function operations: (f + g)(x) = f(x) + g(x), (f - g)(x) = f(x) - g(x)

• Multiplication: (f · g)(x) = f(x) · g(x)

• Composition: (f ∘ g)(x) = f(g(x))

• Evaluation: Replace x with the given value throughout the function

• Order of operations: Always follow PEMDAS when simplifying

Solution: Exercises 4 to 5
4 Domain and Range
Exercise 4
Given h(x) = √(x - 2) + 3, find the domain and range. Then evaluate h(6) and h(11).
Definition:

Domain: All possible x-values for which the function is defined

Range: All possible y-values the function can produce

Function
h(x) = √(x - 2) + 3
Domain
x ≥ 2
Range
y ≥ 3
Step 1: Find the domain

For h(x) = √(x - 2) + 3, the expression under the square root must be non-negative

So: x - 2 ≥ 0

Therefore: x ≥ 2

Domain: [2, ∞)

Step 2: Find the range

Since √(x - 2) ≥ 0 for all x in the domain

Then √(x - 2) + 3 ≥ 3

The smallest value occurs when x = 2: h(2) = √(0) + 3 = 3

As x increases, h(x) increases without bound

Range: [3, ∞)

Step 3: Evaluate h(6)

h(6) = √(6 - 2) + 3 = √4 + 3 = 2 + 3 = 5

Step 4: Evaluate h(11)

h(11) = √(11 - 2) + 3 = √9 + 3 = 3 + 3 = 6

Step 5: Verify domain restrictions

h(6) is valid since 6 ≥ 2 ✓

h(11) is valid since 11 ≥ 2 ✓

h(1) would be invalid since 1 < 2 (negative under square root)

Domain: [2, ∞)
Range: [3, ∞)
h(6) = 5
h(11) = 6
Final answer:

Domain: x ≥ 2 (or [2, ∞))

Range: y ≥ 3 (or [3, ∞))

h(6) = 5

h(11) = 6

Applied rules:

• Square root domain: Expression under √ must be ≥ 0

• Domain finding: Identify restrictions based on function type

• Range finding: Analyze function behavior over its domain

5 Piecewise Functions
Exercise 5
Given f(x) = { 2x + 1 if x < 0, x² if x ≥ 0 }, find f(-3), f(0), and f(4). Also state the domain.
Definition:

Piecewise function: A function defined by different expressions over different intervals

Domain consideration: Each piece has its own domain condition

Piecewise Function
f(x) = {2x + 1 if x < 0, x² if x ≥ 0}
Evaluations
f(-3) = -5, f(0) = 0, f(4) = 16
Step 1: Understand the function definition

f(x) = { 2x + 1 if x < 0

{ x² if x ≥ 0

This means use 2x + 1 when x is negative, use x² when x is non-negative

Step 2: Evaluate f(-3)

Since -3 < 0, use the first piece: f(x) = 2x + 1

f(-3) = 2(-3) + 1 = -6 + 1 = -5

Step 3: Evaluate f(0)

Since 0 ≥ 0, use the second piece: f(x) = x²

f(0) = 0² = 0

Step 4: Evaluate f(4)

Since 4 ≥ 0, use the second piece: f(x) = x²

f(4) = 4² = 16

Step 5: Determine the domain

All real numbers are covered by the conditions x < 0 and x ≥ 0

Domain: (-∞, ∞) or all real numbers

Step 6: Verify evaluations

f(-3) = -5: Since -3 < 0, use 2(-3) + 1 = -5 ✓

f(0) = 0: Since 0 ≥ 0, use 0² = 0 ✓

f(4) = 16: Since 4 ≥ 0, use 4² = 16 ✓

f(-3) = -5
f(0) = 0
f(4) = 16
Domain: All real numbers
Final answer:

f(-3) = -5

f(0) = 0

f(4) = 16

Domain: All real numbers (-∞, ∞)

Applied rules:

• Piecewise evaluation: Determine which condition the input satisfies

• Condition checking: Carefully compare input to condition boundaries

• Domain coverage: Ensure all possible inputs are addressed by conditions

Comprehensive Guide: Function Notation and Applications
f: X → Y
Function Mapping
Key definitions:

Function: A relation where each element in the domain maps to exactly one element in the range

Function notation: f(x) indicates that f is a function of x

Input/Output: x is the input (independent variable), f(x) is the output (dependent variable)

Comprehensive problem-solving approach:
  1. Identify the function: Recognize the function notation and rule
  2. Understand the operation: Determine if evaluating, combining, or composing
  3. Apply the method: Use appropriate function operation rules
  4. Substitute carefully: Replace variables with given values
  5. Simplify systematically: Follow order of operations
  6. Verify results: Check against domain restrictions and reasonableness
Tip 1: Remember f(x) doesn't mean f times x, it means "f of x" - the output of function f when x is the input.
Tip 2: For composite functions, always work from inside out: f(g(x)) means "apply g first, then f".
Tip 3: When dealing with radicals, denominators, or logarithms, always check domain restrictions first.
Tip 4: Practice with different types of functions (linear, quadratic, radical, rational) to build fluency.
Common misconceptions: Confusing f(x) with multiplication, forgetting to apply operations to entire expressions, not considering domain restrictions.
Memory aids: f(x) = "function of x", composition order: innermost function evaluated first.
Essential formulas and relationships:

• Function evaluation: f(a) means substitute a for every x in f(x)

• Arithmetic operations: (f ± g)(x) = f(x) ± g(x), (fg)(x) = f(x)·g(x)

• Division: (f/g)(x) = f(x)/g(x), where g(x) ≠ 0

• Composition: (f ∘ g)(x) = f(g(x))

• Domain restrictions: √(expression) → expression ≥ 0, 1/(expression) → expression ≠ 0

Visualization: Function Behavior Analysis
Exercise 6: Function Comparison
Compare the functions:
f(x) = 2x + 1
g(x) = x²
h(x) = √x

Analysis: Different function types exhibit different growth patterns and behaviors.

  • Linear functions grow at constant rate
  • Quadratic functions grow quadratically
  • Square root functions grow slowly

Questions & Answers

Question: I'm confused about f(x). Is it the same as y? Why do we use f(x) instead of just y?

Answer: Great question! f(x) and y are related but serve different purposes:

f(x) vs y:

  • y is just a variable that represents output values
  • f(x) explicitly shows that y depends on x through function f
  • f(x) allows us to work with multiple functions simultaneously (f(x), g(x), h(x))

For example, if f(x) = 2x + 1, then y = 2x + 1. But if we also have g(x) = x², we can distinguish between f(x) and g(x) easily.

f(x) notation emphasizes the input-output relationship and allows for more sophisticated mathematical operations like function composition.

Think of f(x) as "the output of function f when the input is x."

Question: When evaluating f(a + h), do I substitute (a + h) for every x in the function?

Answer: Yes, absolutely! You must substitute the entire expression (a + h) for every occurrence of x in the function.

For example, if f(x) = x² + 3x + 2, then:

f(a + h) = (a + h)² + 3(a + h) + 2

= a² + 2ah + h² + 3a + 3h + 2

Notice how (a + h) replaced every x, and we had to apply the distributive property and expand (a + h)².

This concept is crucial for calculus (difference quotients) and advanced algebra. Always substitute the complete expression in parentheses to avoid errors!

Common mistake: Writing f(a + h) = a² + h² + 3a + 3h + 2 (incorrectly distributing the square).

Question: How do I remember the order for function composition? Is (f ∘ g)(x) the same as (g ∘ f)(x)?

Answer: The order is crucial! (f ∘ g)(x) means "f of g of x," and you work from inside out:

  • (f ∘ g)(x) = f(g(x)): Apply g first, then apply f to the result
  • (g ∘ f)(x) = g(f(x)): Apply f first, then apply g to the result

Memory aid: Think of it as reading from right to left - the rightmost function is applied first.

No, (f ∘ g)(x) is generally NOT the same as (g ∘ f)(x). Composition is not commutative!

Example: If f(x) = 2x and g(x) = x + 1

  • (f ∘ g)(x) = f(g(x)) = f(x + 1) = 2(x + 1) = 2x + 2
  • (g ∘ f)(x) = g(f(x)) = g(2x) = 2x + 1

Notice: 2x + 2 ≠ 2x + 1, so order matters!

Question: How do I find the domain of a function like f(x) = 1/(x - 3) or g(x) = √(x + 2)?

Answer: Finding domains involves identifying restrictions that would make the function undefined:

For f(x) = 1/(x - 3):

  • Denominators cannot equal zero
  • So x - 3 ≠ 0, which means x ≠ 3
  • Domain: All real numbers except 3, written as (-∞, 3) ∪ (3, ∞)

For g(x) = √(x + 2):

  • Expressions under square roots must be non-negative
  • So x + 2 ≥ 0, which means x ≥ -2
  • Domain: [-2, ∞)

Common domain restrictions:

  • Rational functions: Denominator ≠ 0
  • Radical functions: Expression under √ ≥ 0
  • Logarithmic functions: Argument > 0

Always check what values would break the function!

Question: How do I evaluate piecewise functions? What if x equals the boundary value?

Answer: Evaluating piecewise functions requires careful attention to the conditions:

  1. Determine which condition your x-value satisfies
  2. Use the corresponding function rule
  3. Evaluate using that rule

For boundary values, look at the inequality signs:

If f(x) = { x² if x < 3, 2x + 1 if x ≥ 3 }

  • To find f(3): Since 3 ≥ 3 is true, use 2x + 1 rule: f(3) = 2(3) + 1 = 7
  • To find f(2.99): Since 2.99 < 3, use x² rule: f(2.99) = (2.99)²

The boundary value follows the condition that includes equality (≥ or ≤), not the strict inequality (< or >).

Always double-check which piece applies to your specific x-value!