Session 04-05: Tasks

Rational & Logarithmic Functions

Rational & Logarithmic Functions

Problem 1: Basic Asymptote Identification (x)

For each rational function, identify all vertical and horizontal asymptotes:

  1. \(f(x) = \frac{3}{x-2}\)

  2. \(g(x) = \frac{2x+1}{x-3}\)

  3. \(h(x) = \frac{x^2+1}{x^2-4}\)

  4. \(k(x) = \frac{x^3+2x}{x^2+1}\)

Part a: \(f(x) = \frac{3}{x-2}\)

Step 1: Find vertical asymptotes

  • Set denominator = 0: \(x - 2 = 0\)
  • Vertical asymptote: \(x = 2\)

Step 2: Find horizontal asymptote

  • Degree of numerator: 0
  • Degree of denominator: 1
  • Since deg(num) < deg(den): Horizontal asymptote at \(y = 0\)

Part b: \(g(x) = \frac{2x+1}{x-3}\)

Step 1: Find vertical asymptotes

  • Set denominator = 0: \(x - 3 = 0\)
  • Vertical asymptote: \(x = 3\)

Step 2: Find horizontal asymptote

  • Degree of numerator: 1
  • Degree of denominator: 1
  • Since deg(num) = deg(den): \(y = \frac{2}{1} = 2\)

Part c: \(h(x) = \frac{x^2+1}{x^2-4}\)

Step 1: Find vertical asymptotes

  • Set denominator = 0: \(x^2 - 4 = 0\)
  • \((x-2)(x+2) = 0\)
  • Vertical asymptotes: \(x = 2\) and \(x = -2\)

Step 2: Find horizontal asymptote

  • Degree of numerator: 2
  • Degree of denominator: 2
  • Since deg(num) = deg(den): \(y = \frac{1}{1} = 1\)

Part d: \(k(x) = \frac{x^3+2x}{x^2+1}\)

Step 1: Find vertical asymptotes

  • Set denominator = 0: \(x^2 + 1 = 0\)
  • \(x^2 = -1\) (no real solutions)
  • No vertical asymptotes

Step 2: Find horizontal asymptote

  • Degree of numerator: 3
  • Degree of denominator: 2
  • Since deg(num) > deg(den): No horizontal asymptote
  • (Has oblique asymptote instead)

Problem 2: Holes vs Asymptotes (x)

Identify whether each function has a hole or vertical asymptote at the given point:

  1. \(f(x) = \frac{x^2-9}{x-3}\) at \(x = 3\)

  2. \(g(x) = \frac{x^2-4x+4}{x-2}\) at \(x = 2\)

  3. \(h(x) = \frac{x^2-1}{(x-1)(x+2)}\) at \(x = 1\)

Part a: \(f(x) = \frac{x^2-9}{x-3}\) at \(x = 3\)

Step 1: Factor numerator \[f(x) = \frac{(x-3)(x+3)}{x-3}\]

Step 2: Identify feature

  • Common factor \((x-3)\) cancels
  • This creates a hole at \(x = 3\)

Step 3: Find y-coordinate of hole

  • Simplified function: \(f(x) = x + 3\) for \(x \neq 3\)
  • At \(x = 3\): \(y = 3 + 3 = 6\)
  • Hole at \((3, 6)\)

Part b: \(g(x) = \frac{x^2-4x+4}{x-2}\) at \(x = 2\)

Step 1: Factor numerator \[g(x) = \frac{(x-2)^2}{x-2}\]

Step 2: Simplify \[g(x) = x - 2 \text{ for } x \neq 2\]

Step 3: Identify feature

  • Common factor cancels
  • Hole at \((2, 0)\)

Part c: \(h(x) = \frac{x^2-1}{(x-1)(x+2)}\) at \(x = 1\)

Step 1: Factor numerator \[h(x) = \frac{(x-1)(x+1)}{(x-1)(x+2)}\]

Step 2: Simplify \[h(x) = \frac{x+1}{x+2} \text{ for } x \neq 1\]

Step 3: Find hole coordinate

  • At \(x = 1\): \(y = \frac{1+1}{1+2} = \frac{2}{3}\)
  • Hole at \((1, \frac{2}{3})\)

Problem 3: Average Cost Analysis (xx)

A company has total cost function \(C(x) = 3600 + 24x + 0.01x^2\) dollars for producing \(x\) units.

  1. Find the average cost function \(AC(x)\)
  2. Find the production level that minimizes average cost
  3. What is the minimum average cost?
  4. Find \(\lim_{x \to \infty} AC(x)\) and interpret its meaning

Part a: Average cost function

Step 1: Use formula \(AC(x) = \frac{C(x)}{x}\) \[AC(x) = \frac{3600 + 24x + 0.01x^2}{x}\]

Step 2: Simplify \[AC(x) = \frac{3600}{x} + 24 + 0.01x\]

Part b: Minimize average cost

Step 1: Find derivative of \(AC(x)\) \[AC'(x) = -\frac{3600}{x^2} + 0.01\]

Step 2: Set derivative equal to zero \[-\frac{3600}{x^2} + 0.01 = 0\] \[\frac{3600}{x^2} = 0.01\] \[x^2 = \frac{3600}{0.01} = 360000\] \[x = 600 \text{ units}\]

Part c: Minimum average cost

Step 1: Evaluate \(AC(600)\) \[AC(600) = \frac{3600}{600} + 24 + 0.01(600)\] \[AC(600) = 6 + 24 + 6 = \$36 \text{ per unit}\]

Part d: Limit as x approaches infinity

Step 1: Evaluate limit \[\lim_{x \to \infty} AC(x) = \lim_{x \to \infty} \left(\frac{3600}{x} + 24 + 0.01x\right)\] \[= 0 + 24 + \infty = \infty\]

Step 2: Interpretation As production increases indefinitely, the average cost approaches infinity due to the quadratic term. This represents increasing marginal costs at high production levels (diminishing returns to scale).

Problem 4: Logarithmic Properties (xx)

Simplify each expression using logarithm properties:

  1. \(\log_3(27x^2)\)

  2. \(\ln(e^{2x} \cdot \sqrt{x})\)

  3. \(2\log_5(5x) - \log_5(x^2)\)

  4. \(\log_2(8) + \log_2(x/4)\)

Part a: \(\log_3(27x^2)\)

Step 1: Use product rule \[\log_3(27x^2) = \log_3(27) + \log_3(x^2)\]

Step 2: Evaluate and use power rule \[= \log_3(3^3) + 2\log_3(x)\] \[= 3 + 2\log_3(x)\]

Part b: \(\ln(e^{2x} \cdot \sqrt{x})\)

Step 1: Use product rule \[\ln(e^{2x} \cdot \sqrt{x}) = \ln(e^{2x}) + \ln(\sqrt{x})\]

Step 2: Simplify \[= 2x + \ln(x^{1/2})\] \[= 2x + \frac{1}{2}\ln(x)\]

Part c: \(2\log_5(5x) - \log_5(x^2)\)

Step 1: Expand first term \[2\log_5(5x) = 2[\log_5(5) + \log_5(x)]\] \[= 2[1 + \log_5(x)] = 2 + 2\log_5(x)\]

Step 2: Expand second term \[\log_5(x^2) = 2\log_5(x)\]

Step 3: Combine \[2 + 2\log_5(x) - 2\log_5(x) = 2\]

Part d: \(\log_2(8) + \log_2(x/4)\)

Step 1: Evaluate \(\log_2(8)\) \[\log_2(8) = \log_2(2^3) = 3\]

Step 2: Expand second term \[\log_2(x/4) = \log_2(x) - \log_2(4) = \log_2(x) - 2\]

Step 3: Combine \[3 + \log_2(x) - 2 = 1 + \log_2(x)\]

Problem 5: Graph Sketching (xx)

Sketch the rational function \(f(x) = \frac{2x-4}{x+1}\) by finding:

  1. Domain and asymptotes
  2. x and y intercepts
  3. Sign analysis
  4. End behavior

Part a: Domain and asymptotes

Domain:

  • Undefined when \(x + 1 = 0\)
  • Domain: all real numbers except \(x = -1\)

Vertical asymptote:

  • At \(x = -1\) (where denominator = 0)

Horizontal asymptote:

  • deg(num) = deg(den) = 1
  • \(y = \frac{2}{1} = 2\)

Part b: Intercepts

x-intercept:

  • Set \(f(x) = 0\): \(\frac{2x-4}{x+1} = 0\)
  • Numerator = 0: \(2x - 4 = 0\)
  • \(x = 2\)
  • x-intercept: \((2, 0)\)

y-intercept:

  • Set \(x = 0\): \(f(0) = \frac{2(0)-4}{0+1} = -4\)
  • y-intercept: \((0, -4)\)

Part c: Sign analysis

Interval \((-\infty, -1)\) \((-1, 2)\) \((2, \infty)\)
Test point \(x = -2\) \(x = 0\) \(x = 3\)
\(2x - 4\) \(-8\) (negative) \(-4\) (negative) \(2\) (positive)
\(x + 1\) \(-1\) (negative) \(1\) (positive) \(4\) (positive)
\(f(x)\) positive negative positive

Part d: End behavior

As \(x \to +\infty\): \(f(x) \to 2\) from below As \(x \to -\infty\): \(f(x) \to 2\) from above

Graph features:

  • Approaches \(y = 2\) horizontally
  • Vertical asymptote at \(x = -1\)
  • Changes from positive to negative at asymptote
  • Crosses x-axis at \(x = 2\)

Problem 6: Logarithmic Equation (xxx)

Solve the equation: \(\log_3(x+8) + \log_3(x) = 2\)

Step 1: Combine logs on left side using product rule \[\log_3[x(x+8)] = 2\]

Step 2: Convert to exponential form \[x(x+8) = 3^2 = 9\]

Step 3: Expand left side \[x^2 + 8x = 9\]

Step 4: Rearrange to standard form \[x^2 + 8x - 9 = 0\]

Step 5: Factor the quadratic We need two numbers that multiply to -9 and add to 8. These are 9 and -1. \[(x + 9)(x - 1) = 0\]

Step 6: Solve for x \[x + 9 = 0 \text{ or } x - 1 = 0\] \[x = -9 \text{ or } x = 1\]

Step 7: Check domain restrictions For logarithms to be defined:

  • \(x + 8 > 0 \Rightarrow x > -8\)
  • \(x > 0\)

Most restrictive: \(x > 0\)

Therefore, \(x = -9\) is rejected (doesn’t satisfy \(x > 0\))

Step 8: Final answer \[x = 1\]

Verification: \[\log_3(1+8) + \log_3(1) = \log_3(9) + \log_3(1) = 2 + 0 = 2\] ✓

Problem 7: Semi-log Data Analysis (xxx)

A bacteria culture shows the following population data:

Time (hours) 0 2 4 6 8
Population 100 400 1600 6400 25600
  1. Show that this represents exponential growth
  2. Find the growth formula \(P(t) = P_0 \cdot b^t\)
  3. What is the doubling time?
  4. Predict the population at \(t = 10\) hours

Part a: Show exponential growth

Step 1: Check ratios between consecutive terms

  • \(\frac{P(2)}{P(0)} = \frac{400}{100} = 4\)
  • \(\frac{P(4)}{P(2)} = \frac{1600}{400} = 4\)
  • \(\frac{P(6)}{P(4)} = \frac{6400}{1600} = 4\)
  • \(\frac{P(8)}{P(6)} = \frac{25600}{6400} = 4\)

Step 2: Constant ratio confirms exponential growth

  • Each 2-hour period multiplies population by 4
  • This is characteristic of exponential growth

Part b: Find growth formula

Step 1: Identify initial value \(P_0 = 100\) (at \(t = 0\))

Step 2: Find base from 2-hour growth After 2 hours: \(100 \cdot b^2 = 400\) \[b^2 = 4\] \[b = 2\]

Step 3: Write formula \[P(t) = 100 \cdot 2^t\]

Part c: Doubling time

Step 1: Find when \(P(t) = 2 \cdot P_0 = 200\) \[100 \cdot 2^t = 200\] \[2^t = 2\] \[t = 1 \text{ hour}\]

Doubling time: 1 hour

Part d: Population at t = 10

Step 1: Use formula \[P(10) = 100 \cdot 2^{10}\]

Step 2: Calculate \[P(10) = 100 \cdot 1024 = 102,400\]

Answer: 102,400 bacteria after 10 hours

Problem 8: Complex Rational Function (xxxx)

Analyze the function \(f(x) = \frac{x^2 - 4}{x - 1}\) completely:

  1. Find all asymptotes and holes
  2. Find the x and y intercepts
  3. Find where \(f(x) = x + 2\)
  4. Sketch the function showing all key features

Part a: Asymptotes and holes

Step 1: Factor numerator if possible \[x^2 - 4 = (x-2)(x+2)\]

Step 2: Check for common factors with denominator

  • Numerator: \((x-2)(x+2)\)
  • Denominator: \((x-1)\)
  • No common factors, so no holes

Step 3: Find vertical asymptote

  • Denominator = 0 when \(x = 1\)
  • Vertical asymptote: \(x = 1\)

Step 4: Check horizontal asymptote

  • deg(numerator) = 2, deg(denominator) = 1
  • Since deg(num) > deg(den), no horizontal asymptote

Part b: Intercepts

x-intercepts:

  • Set numerator = 0: \((x-2)(x+2) = 0\)
  • Solutions: \(x = 2\) and \(x = -2\)
  • x-intercepts: \((2, 0)\) and \((-2, 0)\)

y-intercept:

  • Evaluate \(f(0) = \frac{0^2 - 4}{0 - 1} = \frac{-4}{-1} = 4\)
  • y-intercept: \((0, 4)\)

Part c: Find where f(x) = x + 2

Step 1: Set up equation \[\frac{x^2 - 4}{x - 1} = x + 2\]

Step 2: Multiply both sides by \((x - 1)\), assuming \(x \neq 1\) \[x^2 - 4 = (x + 2)(x - 1)\]

Step 3: Expand right side \[x^2 - 4 = x^2 + 2x - x - 2\] \[x^2 - 4 = x^2 + x - 2\]

Step 4: Simplify \[-4 = x - 2\] \[x = -2\]

Step 5: Verify \[f(-2) = \frac{(-2)^2 - 4}{-2 - 1} = \frac{4 - 4}{-3} = 0\] \[x + 2 = -2 + 2 = 0\] ✓

Solution: \(x = -2\)

Part d: Key features for sketch

Summary of features:

  • Domain: all real except \(x = 1\)
  • Vertical asymptote: \(x = 1\)
  • Oblique asymptote: \(y = x + 1\)
  • x-intercepts: \((2, 0)\) and \((-2, 0)\)
  • y-intercept: \((0, 4)\)
  • Point where \(f(x) = x + 2\): \((-2, 0)\)
  • As \(x \to \infty\): \(f(x) \approx x + 1\) (from above)
  • As \(x \to -\infty\): \(f(x) \approx x + 1\) (from below)