Solved Exercises on Evaluating Functions in Integrated Math 1

Master evaluating functions: substitution, composition, and real-world applications through these 5 detailed exercises.

Solution: Exercises 1 to 3
1 Basic function evaluation
Exercise 1
Given f(x) = 3x - 4, evaluate f(5), f(-2), and f(0).
Definition:

Function evaluation: Replacing the input variable with a specific value to find the corresponding output.

Substitution: The process of replacing variables with numbers.

Evaluation method:
  1. Replace the variable x with the given value
  2. Perform arithmetic operations following order of operations
  3. Simplify to get the final result
Function
f(x) = 3x - 4
Evaluations
f(5)=11, f(-2)=-10, f(0)=-4
Step 1: Evaluate f(5)

Replace x with 5: f(5) = 3(5) - 4 = 15 - 4 = 11

Step 2: Evaluate f(-2)

Replace x with -2: f(-2) = 3(-2) - 4 = -6 - 4 = -10

Step 3: Evaluate f(0)

Replace x with 0: f(0) = 3(0) - 4 = 0 - 4 = -4

f(5) = 11, f(-2) = -10, f(0) = -4
Final answer:

f(5) = 11, f(-2) = -10, f(0) = -4

These values represent the output of the function when the input is 5, -2, and 0 respectively.

Applied rules:

• Substitution: Replace variable with given value

• Order of operations: Parentheses, exponents, multiplication/division, addition/subtraction

• Arithmetic: Perform calculations accurately

2 Function evaluation with expressions
Exercise 2
Given g(x) = x² + 3x - 2, evaluate g(a), g(2a), and g(x + 1).
Definition:

Variable substitution: Replacing the input variable with another variable or expression.

Algebraic expression: A combination of variables, numbers, and operations.

Function
g(x) = x² + 3x - 2
Evaluations
g(a)=a²+3a-2, g(2a)=4a²+6a-2, g(x+1)=x²+5x+2
Step 1: Evaluate g(a)

Replace x with a: g(a) = a² + 3a - 2

Step 2: Evaluate g(2a)

Replace x with 2a: g(2a) = (2a)² + 3(2a) - 2 = 4a² + 6a - 2

Step 3: Evaluate g(x + 1)

Replace x with (x + 1): g(x + 1) = (x + 1)² + 3(x + 1) - 2

= x² + 2x + 1 + 3x + 3 - 2 = x² + 5x + 2

g(a) = a² + 3a - 2
Final answer:

g(a) = a² + 3a - 2

g(2a) = 4a² + 6a - 2

g(x + 1) = x² + 5x + 2

Applied rules:

• Substitution: Replace variable with expression

• Distribution: Apply multiplication to each term in parentheses

• Expansion: Expand squared terms using (a+b)² = a² + 2ab + b²

3 Function evaluation with rational expressions
Exercise 3
Given h(x) = (2x + 1)/(x - 3), evaluate h(4) and h(-1). State any restrictions.
Definition:

Rational function: A function expressed as a ratio of two polynomials.

Domain restriction: Values that make the denominator zero are excluded.

Function
h(x) = (2x + 1)/(x - 3)
Evaluations
h(4)=9, h(-1)=-1/4
Restriction
x ≠ 3
Step 1: Identify restrictions

Denominator cannot be zero: x - 3 ≠ 0, so x ≠ 3

Step 2: Evaluate h(4)

h(4) = (2(4) + 1)/(4 - 3) = (8 + 1)/1 = 9/1 = 9

Step 3: Evaluate h(-1)

h(-1) = (2(-1) + 1)/((-1) - 3) = (-2 + 1)/(-4) = -1/(-4) = 1/4

h(4) = 9, h(-1) = 1/4
Final answer:

h(4) = 9, h(-1) = 1/4

Domain restriction: x ≠ 3 (since this would make the denominator zero)

Applied rules:

• Domain restrictions: Exclude values that make denominator zero

• Rational arithmetic: Perform operations in numerator and denominator separately

• Sign rules: Negative divided by negative equals positive

Evaluating Functions Guide
\(f(a) = \text{result when } x = a\)
Function Evaluation
Key definitions:

Function evaluation: The process of finding the output value of a function for a given input value

Substitution: Replacing the variable in a function with a specific value or expression

Domain restriction: Values that must be excluded from the domain due to mathematical impossibilities

Complete methodology:
  1. Identify the function: Note the function rule and variable
  2. Identify the input: Determine the value or expression to substitute
  3. Substitute: Replace the variable with the given input
  4. Calculate: Perform arithmetic operations following order of operations
  5. Simplify: Reduce to simplest form
Tip 1: Always use parentheses when substituting negative numbers to avoid sign errors
Tip 2: Check that the input value is in the domain of the function
Tip 3: When substituting expressions, apply the function rule to the entire expression
Tip 4: Be careful with order of operations, especially with exponents and division
Common errors: Forgetting parentheses with negative numbers, making arithmetic mistakes, ignoring domain restrictions
Exam preparation: Practice with various function types (linear, quadratic, rational), work with algebraic expressions as inputs
Formulas to know by heart:

• Function evaluation: f(a) means replace x with a in f(x)

• Order of operations: PEMDAS (Parentheses, Exponents, Multiplication/Division, Addition/Subtraction)

• Domain restrictions: Exclude values that cause division by zero or negative under even roots

Solution: Exercises 4 to 5
4 Function composition evaluation
Exercise 4
Given f(x) = 2x - 1 and g(x) = x², find f(g(3)) and g(f(3)).
Definition:

Function composition: Applying one function to the result of another function.

Notation: f(g(x)) means "f of g of x" or g(x) is the input for f.

Functions
f(x) = 2x - 1, g(x) = x²
Compositions
f(g(3)) = 17, g(f(3)) = 25
Step 1: Find f(g(3))

First evaluate g(3): g(3) = 3² = 9

Then evaluate f(9): f(9) = 2(9) - 1 = 18 - 1 = 17

So f(g(3)) = 17

Step 2: Find g(f(3))

First evaluate f(3): f(3) = 2(3) - 1 = 6 - 1 = 5

Then evaluate g(5): g(5) = 5² = 25

So g(f(3)) = 25

f(g(3)) = 17, g(f(3)) = 25
Final answer:

f(g(3)) = 17 and g(f(3)) = 25

Note: Function composition is not commutative - f(g(x)) ≠ g(f(x)) in general.

Applied rules:

• Composition: Work from inside to outside

• Order matters: f(g(x)) ≠ g(f(x)) in general

• Sequential evaluation: Evaluate inner function first

5 Real-world function evaluation
Exercise 5
The population P (in thousands) of a city after t years is given by P(t) = 50 + 3t - 0.1t². Find P(0), P(5), and P(10). Interpret the meaning of P(0).
Definition:

Real-world function: A function that models a practical situation with meaningful inputs and outputs.

Initial value: The function value when the input is zero.

Function
P(t) = 50 + 3t - 0.1t²
Evaluations
P(0)=50, P(5)=62.5, P(10)=70
Step 1: Find P(0)

P(0) = 50 + 3(0) - 0.1(0)² = 50 + 0 - 0 = 50

Step 2: Find P(5)

P(5) = 50 + 3(5) - 0.1(5)² = 50 + 15 - 0.1(25) = 50 + 15 - 2.5 = 62.5

Step 3: Find P(10)

P(10) = 50 + 3(10) - 0.1(10)² = 50 + 30 - 0.1(100) = 50 + 30 - 10 = 70

Step 4: Interpret P(0)

P(0) = 50 represents the initial population of the city (at time t = 0), which is 50 thousand people.

P(0) = 50 thousand
Final answer:

P(0) = 50 (initial population: 50,000 people)

P(5) = 62.5 (population after 5 years: 62,500 people)

P(10) = 70 (population after 10 years: 70,000 people)

P(0) represents the initial population of the city at the starting time.

Applied rules:

• Real-world context: Connect mathematical results to practical meaning

• Units: Include appropriate units in the answer

• Interpretation: Explain what the function values represent

Evaluating Functions: Complete Summary
\(f(a) = \text{result when } x = a\)
Function Evaluation
Key definitions:

Function evaluation: The process of finding the output value of a function for a given input value

Substitution: Replacing the variable in a function with a specific value or expression

Domain restriction: Values that must be excluded from the domain due to mathematical impossibilities

Function composition: Applying one function to the result of another function

Initial value: The function value when the input is zero

Complete methodology:
  1. Identify the function: Note the function rule and variable
  2. Identify the input: Determine the value or expression to substitute
  3. Substitute: Replace the variable with the given input
  4. Calculate: Perform arithmetic operations following order of operations
  5. Simplify: Reduce to simplest form
  6. Interpret: Connect results to real-world meaning when applicable
Tip 1: Always use parentheses when substituting negative numbers to avoid sign errors
Tip 2: Check that the input value is in the domain of the function
Tip 3: When substituting expressions, apply the function rule to the entire expression
Tip 4: Be careful with order of operations, especially with exponents and division
Tip 5: For function composition, work from inside to outside: f(g(x)) means evaluate g(x) first, then use that result as input for f
Common errors: Forgetting parentheses with negative numbers, making arithmetic mistakes, ignoring domain restrictions
Exam preparation: Practice with various function types (linear, quadratic, rational), work with algebraic expressions as inputs
Formulas to know by heart:

• Function evaluation: f(a) means replace x with a in f(x)

• Order of operations: PEMDAS (Parentheses, Exponents, Multiplication/Division, Addition/Subtraction)

• Domain restrictions: Exclude values that cause division by zero or negative under even roots

• Composition: f(g(x)) means g(x) is input for f

Function Evaluation Patterns
Evaluation Analysis: Quadratic Function
Consider the function f(x) = x² - 4x + 3:
• Evaluate f(-1), f(0), f(1), f(2), f(3), f(4), f(5)
• Observe the pattern of outputs
• Notice the symmetry around x = 2
This demonstrates how function evaluation reveals patterns.

Analysis: Function evaluation helps identify key features like symmetry, minimum/maximum values, and intercepts.

  • Evaluation at integer values reveals the function's behavior
  • Symmetry occurs around the vertex of the parabola
  • Patterns in outputs can indicate function type

Questions & Answers

Question: I get confused when evaluating functions with exponents, like f(x) = x² + 2x - 1 and I need to find f(-3). Do I square the negative number?

Answer: Yes, you do square the negative number! Here's the correct process:

f(x) = x² + 2x - 1, find f(-3)

  • Substitute: f(-3) = (-3)² + 2(-3) - 1
  • Calculate each term:
    • (-3)² = 9 (negative squared is positive)
    • 2(-3) = -6 (positive times negative is negative)
    • Last term: -1
  • Combine: f(-3) = 9 + (-6) + (-1) = 9 - 6 - 1 = 2

Key tip: Always use parentheses when substituting negative values to keep track of signs correctly.

Question: What's the difference between f(g(x)) and f(x)·g(x)? They look similar but I know they're different.

Answer: These are completely different operations:

  • Function composition f(g(x)): Apply g to x first, then apply f to the result of g(x)
  • Function multiplication f(x)·g(x): Multiply the outputs of f(x) and g(x) together

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

  • f(g(x)) = f(2x) = (2x) + 1 = 2x + 1
  • f(x)·g(x) = (x + 1)(2x) = 2x² + 2x

Notice how these produce completely different results!

Question: How do I handle function evaluation when the function has multiple terms with different operations?

Answer: Always follow the order of operations (PEMDAS/BODMAS) when evaluating functions:

  1. Parentheses: Handle anything in parentheses first
  2. Exponents: Calculate all powers and roots
  3. Multiplication/Division: From left to right
  4. Addition/Subtraction: From left to right

Example: For f(x) = 2x² - 3(x + 1) + 4, evaluate f(2)

  • f(2) = 2(2)² - 3(2 + 1) + 4
  • = 2(4) - 3(3) + 4
  • = 8 - 9 + 4 = 3

Work step by step, following the order of operations at each stage.