MathIsimple
Lesson 1.3: Functions vs Inequalities & Model Validation

Master Function-Inequality Connections & Model Validation

Bridge the gap between functions and inequalities! Learn to solve function inequalities, compare function values graphically, validate mathematical models with real data, and interpret solutions in practical contexts. Develop critical thinking skills for model evaluation.

Learning Objectives

Solve function inequalities using algebraic and graphical methods
Compare two functions and find intersection points
Validate mathematical models with real-world data
Interpret solutions in practical contexts
Assess model accuracy and limitations
Apply to real-world decision making

Core Concepts & Theoretical Foundation

Function-Inequality Equivalence

Fundamental Connection: Solving f(x)>g(x)f(x) > g(x) is equivalent to finding where the graph of ff is above the graph of gg.

Single function inequality: f(x)>cf(x) > c where c is a constant

Two function comparison: f(x)>g(x)f(x) > g(x) for all x in domain

Interval solutions: Express solutions as intervals where inequality holds

Boundary points: Points where f(x)=g(x)f(x) = g(x) (intersection points)

Model Validation Framework

Validation Process:

1. Data Collection: Gather real-world data points

2. Model Fitting: Determine model parameters from data

3. Prediction Testing: Use model to predict new values

4. Error Analysis: Compare predictions with actual values

5. Model Refinement: Adjust model based on validation results

Graphical Analysis Techniques

Visual Problem-Solving:

Intersection method: Find where graphs cross to determine boundary points

Test point method: Choose points in different regions to determine solution sets

Slope analysis: Use function slopes to predict behavior

Domain restrictions: Consider practical limitations on variable values

Detailed Worked Examples

Example 1: Single Function Inequality

Solve the inequality: 2x1>32x - 1 > 3 using both algebraic and graphical methods.

Algebraic Method:

2x1>32x - 1 > 3

2x>3+12x > 3 + 1

2x>42x > 4

x>2x > 2

Graphical Method:

1. Graph y=2x1y = 2x - 1 (line with slope 2, y-intercept -1)

2. Graph y=3y = 3 (horizontal line at y = 3)

3. Find intersection: 2x1=32x - 1 = 3x=2x = 2

4. Since slope is positive (2 > 0), the line rises from left to right

5. For x>2x > 2, the line y=2x1y = 2x - 1 is above y=3y = 3

Solution: x>2x > 2 or (2,)(2, \infty) in interval notation

Key Insight: For linear functions f(x)=mx+bf(x) = mx + b, the solution to f(x)>cf(x) > cdepends on the sign of the slope m. If m > 0, solution is x>cbmx > \frac{c-b}{m}; if m < 0, solution is x<cbmx < \frac{c-b}{m}.

Example 2: Two Function Comparison

Let f(x)=2x3f(x) = 2x - 3 and g(x)=x+6g(x) = -x + 6. Find all values of x for which f(x)g(x)f(x) \leq g(x).

Method 1: Algebraic Solution

f(x)g(x)f(x) \leq g(x)

2x3x+62x - 3 \leq -x + 6

2x+x6+32x + x \leq 6 + 3

3x93x \leq 9

x3x \leq 3

Method 2: Graphical Analysis

1. Find intersection point: 2x3=x+62x - 3 = -x + 6x=3x = 3

2. At intersection: f(3)=g(3)=3f(3) = g(3) = 3, so point is (3, 3)

3. Analyze slopes: ff has slope 2 (rising), gg has slope -1 (falling)

4. For x<3x < 3: f(x)<g(x)f(x) < g(x) (f is below g)

5. For x>3x > 3: f(x)>g(x)f(x) > g(x) (f is above g)

6. For x=3x = 3: f(x)=g(x)f(x) = g(x) (intersection point)

Solution: x3x \leq 3 or (,3](-\infty, 3] in interval notation

Verification: Test x = 2: f(2) = 1, g(2) = 4, so f(2) < g(2) ✓. Test x = 4: f(4) = 5, g(4) = 2, so f(4) > g(4) ✓.

Example 3: Model Validation with Real Data

A phone's battery life follows the model B(t)=15t+100B(t) = -15t + 100 where B is battery percentage and t is hours of use. Real data shows: t=1→85%, t=2→70%, t=3→55%, t=4→40%. Validate the model and find when battery ≥ 20%.

Step 1: Model Predictions vs. Actual Data

Model predictions: B(1) = 85%, B(2) = 70%, B(3) = 55%, B(4) = 40%

Actual data: 85%, 70%, 55%, 40%

Perfect match! Model appears accurate for this data range.

Step 2: Find when battery ≥ 20%

B(t)20B(t) \geq 20

15t+10020-15t + 100 \geq 20

15t20100-15t \geq 20 - 100

15t80-15t \geq -80

t8015=1635.33t \leq \frac{80}{15} = \frac{16}{3} \approx 5.33 hours

Step 3: Validation Testing

Test t = 5: B(5) = -15(5) + 100 = 25% (above 20%)

Test t = 6: B(6) = -15(6) + 100 = 10% (below 20%)

Model predicts battery drops below 20% between 5 and 6 hours.

Step 4: Model Limitations

• Model assumes linear decline (may not hold for very low battery)

• Doesn't account for different usage patterns

• Battery behavior may change with age

• Temperature and other factors not considered

Practical Application: The phone should be charged before 5.33 hours of use to maintain battery above 20%. However, the model should be revalidated with more data points and different usage scenarios.

Example 4: Real-World Decision Making

Two internet plans: Plan A costs $50/month + $1/GB, Plan B costs $30/month + $2/GB. For what data usage is Plan A more economical? Validate with actual usage data.

Step 1: Set up cost functions

Plan A: CA(x)=50+xC_A(x) = 50 + x where x = GB used

Plan B: CB(x)=30+2xC_B(x) = 30 + 2x where x = GB used

Step 2: Find break-even point

CA(x)=CB(x)C_A(x) = C_B(x)

50+x=30+2x50 + x = 30 + 2x

5030=2xx50 - 30 = 2x - x

20=x20 = x

At 20 GB, both plans cost $70.

Step 3: Determine when Plan A is cheaper

CA(x)<CB(x)C_A(x) < C_B(x)

50+x<30+2x50 + x < 30 + 2x

20<x20 < x

Plan A is cheaper when x > 20 GB.

Step 4: Validation with usage data

Monthly usage: 15 GB, 25 GB, 35 GB, 45 GB

At 15 GB: Plan A = $65, Plan B = $60 → Plan B cheaper

At 25 GB: Plan A = $75, Plan B = $80 → Plan A cheaper

At 35 GB: Plan A = $85, Plan B = $100 → Plan A cheaper

At 45 GB: Plan A = $95, Plan B = $120 → Plan A cheaper

Step 5: Practical recommendation

• If usage < 20 GB: Choose Plan B

• If usage > 20 GB: Choose Plan A

• If usage ≈ 20 GB: Either plan (same cost)

Decision Framework: The mathematical model provides a clear decision rule based on usage patterns. Users should track their actual usage over several months to make the optimal choice.

Advanced Techniques & Model Analysis

Error Analysis & Model Accuracy

Quantitative Validation Methods:

Mean Absolute Error (MAE): MAE=1ni=1nyiy^i\text{MAE} = \frac{1}{n}\sum_{i=1}^{n}|y_i - \hat{y}_i|

Root Mean Square Error (RMSE): RMSE=1ni=1n(yiy^i)2\text{RMSE} = \sqrt{\frac{1}{n}\sum_{i=1}^{n}(y_i - \hat{y}_i)^2}

Percentage Error: PE=yiy^iyi×100%\text{PE} = \frac{|y_i - \hat{y}_i|}{y_i} \times 100\%

R-squared: Proportion of variance explained by the model

Piecewise Function Analysis

When functions have different behaviors in different intervals:

Identify breakpoints: Points where function definition changes

Solve each piece separately: Apply appropriate function rule

Check continuity: Ensure function values match at breakpoints

Combine solutions: Union of solution sets from each piece

Sensitivity Analysis

Understanding how changes in parameters affect solutions:

Parameter variation: Test how solutions change with different parameter values

Robustness testing: Check if conclusions hold under uncertainty

Scenario analysis: Consider best-case, worst-case, and most-likely scenarios

Confidence intervals: Range of values within which true solution likely falls

Common Pitfalls & Error Prevention

Pitfall 1: Ignoring Domain Restrictions

Error: Solving function inequalities without considering practical domain limitations.

Solution: Always check if solutions make sense in the given context (e.g., negative time, impossible quantities).

Pitfall 2: Overfitting Models to Data

Error: Creating models that fit training data perfectly but fail on new data.

Solution: Use cross-validation and test models on independent data sets.

Pitfall 3: Extrapolation Beyond Data Range

Error: Using models to predict values outside the range of training data.

Solution: Clearly state model limitations and avoid predictions beyond observed data range.

Pitfall 4: Confusing Correlation with Causation

Error: Assuming that mathematical relationships imply cause-and-effect relationships.

Solution: Distinguish between mathematical models and causal explanations.

Pitfall 5: Not Accounting for Measurement Error

Error: Treating all data points as equally reliable without considering measurement uncertainty.

Solution: Include error bars and uncertainty analysis in model validation.

Comprehensive Practice Problems

Problem 1: Function Inequality

Solve: 3x+22x13x + 2 \leq 2x - 1 and verify graphically.

Show Solution

Algebraic: 3x+22x13x + 2 \leq 2x - 1x3x \leq -3

Graphical: Graph y = 3x + 2 and y = 2x - 1, find intersection at x = -3

Solution: x3x \leq -3 or (,3](-\infty, -3]

Problem 2: Function Comparison

For f(x)=x24f(x) = x^2 - 4 and g(x)=2x1g(x) = 2x - 1, find when f(x)>g(x)f(x) > g(x).

Show Solution

Set up: x24>2x1x^2 - 4 > 2x - 1x22x3>0x^2 - 2x - 3 > 0

Factor: (x3)(x+1)>0(x-3)(x+1) > 0

Solution: x<1x < -1 or x>3x > 3

Problem 3: Model Validation

A car's fuel efficiency model is E(v)=0.02v2+2v+20E(v) = -0.02v^2 + 2v + 20 (mpg at speed v mph). Test data: (30, 38), (50, 50), (70, 38). Validate the model and find optimal speed.

Show Solution

Model predictions: E(30) = 38, E(50) = 50, E(70) = 38

Actual data: 38, 50, 38 - Perfect match!

Optimal speed: Vertex at v = 50 mph (maximum efficiency)

Problem 4: Real-World Application

Two job offers: Job A pays $40,000 + $2,000 per year of experience, Job B pays $35,000 + $3,000 per year of experience. When is Job A better? Create a decision model.

Show Solution

Salary functions: A(x) = 40000 + 2000x, B(x) = 35000 + 3000x

Break-even: 40000 + 2000x = 35000 + 3000x → x = 5 years

Decision rule: Job A better if experience > 5 years