Session 09-01: Tasks

Advanced Functions - Exam Review

Advanced Functions - Exam Review

Problem 1: Logarithm Rules (x)

Simplify the following expressions using logarithm properties:

  1. \(\log_3(27) + \log_3(9)\)
  2. \(\ln(e^5 \cdot e^{-2})\)
  3. \(\log_2(32) - \log_2(4)\)
  4. \(\log_{10}(500) + \log_{10}(2)\)

a) We use the fact that \(27 = 3^3\) and \(9 = 3^2\):

\[\log_3(27) + \log_3(9) = \log_3(3^3) + \log_3(3^2) = 3 + 2 = 5\]

Alternatively, using the product rule: \(\log_3(27 \cdot 9) = \log_3(243) = \log_3(3^5) = 5\).

b) First simplify using the exponent rule for products, then apply the inverse property:

\[\ln(e^5 \cdot e^{-2}) = \ln(e^{5+(-2)}) = \ln(e^3) = 3\]

c) Using the quotient rule \(\log_b(m) - \log_b(n) = \log_b\!\left(\frac{m}{n}\right)\):

\[\log_2(32) - \log_2(4) = \log_2\!\left(\frac{32}{4}\right) = \log_2(8) = \log_2(2^3) = 3\]

d) Using the product rule \(\log_b(m) + \log_b(n) = \log_b(m \cdot n)\):

\[\log_{10}(500) + \log_{10}(2) = \log_{10}(500 \cdot 2) = \log_{10}(1000) = \log_{10}(10^3) = 3\]

Problem 2: Exponential Equations (x)

Solve for \(x\):

  1. \(5^{x+1} = 125\)
  2. \(2^{3x} = 16\)
  3. \(e^{2x-1} = e^5\)

a) Rewrite \(125\) as a power of \(5\):

\[5^{x+1} = 5^3\]

Since the bases are equal, the exponents must be equal:

\[x + 1 = 3 \implies x = 2\]

b) Rewrite \(16\) as a power of \(2\):

\[2^{3x} = 2^4\]

Equating exponents:

\[3x = 4 \implies x = \frac{4}{3}\]

c) Since the bases are both \(e\), we equate the exponents directly:

\[2x - 1 = 5 \implies 2x = 6 \implies x = 3\]

Problem 3: Asymptote Identification (x)

Find all vertical and horizontal asymptotes:

  1. \(f(x) = \frac{4}{x+3}\)
  2. \(g(x) = \frac{3x-1}{x+2}\)
  3. \(h(x) = \frac{x^2+1}{x^2-9}\)
  4. \(k(x) = \frac{2x^3}{x^2-1}\)

a) The denominator is zero when \(x + 3 = 0\), i.e., \(x = -3\).

  • Vertical asymptote: \(x = -3\)
  • Horizontal asymptote: As \(x \to \pm\infty\), \(f(x) \to 0\), so \(y = 0\).

b) The denominator is zero when \(x = -2\).

  • Vertical asymptote: \(x = -2\)
  • Horizontal asymptote: The degrees of numerator and denominator are both 1. We take the ratio of leading coefficients: \(y = \frac{3}{1} = 3\).

c) The denominator factors as \(x^2 - 9 = (x-3)(x+3)\), which is zero at \(x = 3\) and \(x = -3\). Neither cancels with the numerator \(x^2 + 1\).

  • Vertical asymptotes: \(x = 3\) and \(x = -3\)
  • Horizontal asymptote: Degrees are equal (both 2), so \(y = \frac{1}{1} = 1\).

d) The denominator factors as \(x^2 - 1 = (x-1)(x+1)\), zero at \(x = 1\) and \(x = -1\).

  • Vertical asymptotes: \(x = 1\) and \(x = -1\)
  • Horizontal asymptote: The degree of the numerator (3) is greater than the degree of the denominator (2), so there is no horizontal asymptote. Instead, there is an oblique (slant) asymptote. Performing polynomial long division: \(k(x) = 2x + \frac{2x}{x^2 - 1}\), so the oblique asymptote is \(y = 2x\).

Problem 4: Domain and Range (x)

Find the domain and range:

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

a) The natural logarithm requires a strictly positive argument:

\[x - 3 > 0 \implies x > 3\]

  • Domain: \((3, \infty)\)
  • Range: The logarithm can take any real value, so the range is \((-\infty, \infty)\).

b) The exponential function \(e^x\) is defined for all real numbers and always positive:

  • Domain: \((-\infty, \infty)\)
  • Range: Since \(e^x > 0\) for all \(x\), we have \(e^x + 2 > 2\). The range is \((2, \infty)\).

c) We need \(x^2 - 4 \neq 0\), i.e., \(x \neq \pm 2\).

  • Domain: \((-\infty, -2) \cup (-2, 2) \cup (2, \infty)\)
  • Range: Note that \(x^2 - 4 \geq -4\) with equality at \(x = 0\), so \(h(0) = \frac{1}{-4} = -\frac{1}{4}\). As \(x \to \pm 2\), \(h(x) \to \pm\infty\). As \(x \to \pm\infty\), \(h(x) \to 0^+\). Analyzing carefully: for \(|x| > 2\), \(h(x) > 0\) and can be arbitrarily large or approach \(0^+\), giving \((0, \infty)\). For \(|x| < 2\), \(x^2 - 4 < 0\), and \(h(x) \leq -\frac{1}{4}\), giving \((-\infty, -\frac{1}{4}]\). Therefore, the range is \(\left(-\infty, -\frac{1}{4}\right] \cup (0, \infty)\).

Problem 5: Polynomial Factorization and Zeros (xx)

Factor completely and find all real zeros:

  1. \(p(x) = x^3 - 4x^2 + x + 6\)
  2. \(q(x) = x^4 - 13x^2 + 36\)
  3. \(r(x) = 2x^3 + 3x^2 - 8x + 3\)

a) We try possible rational roots \(\pm 1, \pm 2, \pm 3, \pm 6\):

\(p(1) = 1 - 4 + 1 + 6 = 4 \neq 0\)

\(p(-1) = -1 - 4 - 1 + 6 = 0\) ✓

So \((x + 1)\) is a factor. Performing polynomial division:

\[p(x) = (x + 1)(x^2 - 5x + 6) = (x + 1)(x - 2)(x - 3)\]

Zeros: \(x = -1, \; x = 2, \; x = 3\)

b) This is a quadratic in \(x^2\). Let \(u = x^2\):

\[u^2 - 13u + 36 = (u - 4)(u - 9)\]

Substituting back:

\[q(x) = (x^2 - 4)(x^2 - 9) = (x-2)(x+2)(x-3)(x+3)\]

Zeros: \(x = -3, \; x = -2, \; x = 2, \; x = 3\)

c) Try possible rational roots \(\pm 1, \pm 3, \pm \frac{1}{2}, \pm \frac{3}{2}\):

\(r(1) = 2 + 3 - 8 + 3 = 0\) ✓

So \((x - 1)\) is a factor. Performing polynomial division:

\[r(x) = (x - 1)(2x^2 + 5x - 3)\]

Factor the quadratic using the quadratic formula or by inspection:

\[2x^2 + 5x - 3 = (2x - 1)(x + 3)\]

Therefore:

\[r(x) = (x - 1)(2x - 1)(x + 3)\]

Zeros: \(x = 1, \; x = \frac{1}{2}, \; x = -3\)

Problem 6: Complete Rational Function Analysis (xx)

For \(f(x) = \frac{x^2 - 4x + 3}{x^2 - 5x + 6}\):

  1. Factor numerator and denominator
  2. Identify any holes and vertical asymptotes
  3. Find the horizontal asymptote
  4. Find all x-intercepts and the y-intercept
  5. Sketch the graph

a) Factor both:

\[\text{Numerator: } x^2 - 4x + 3 = (x - 1)(x - 3)\]

\[\text{Denominator: } x^2 - 5x + 6 = (x - 2)(x - 3)\]

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

b) The factor \((x - 3)\) appears in both numerator and denominator, so it cancels:

\[f(x) = \frac{x - 1}{x - 2}, \quad x \neq 3\]

  • Hole at \(x = 3\): The function value of the simplified form at \(x = 3\) is \(\frac{3 - 1}{3 - 2} = 2\). So the hole is at \((3, 2)\).
  • Vertical asymptote at \(x = 2\) (the remaining zero of the denominator that does not cancel).

c) In the simplified form \(\frac{x-1}{x-2}\), the degrees of numerator and denominator are both 1. The horizontal asymptote is:

\[y = \frac{1}{1} = 1\]

d) x-intercepts: Set the numerator of the simplified form to zero: \(x - 1 = 0 \implies x = 1\). So the x-intercept is \((1, 0)\).

Note: \(x = 3\) is NOT an x-intercept because the function is undefined there (it is a hole).

y-intercept: \(f(0) = \frac{(0-1)(0-3)}{(0-2)(0-3)} = \frac{3}{6} = \frac{1}{2}\). So the y-intercept is \(\left(0, \frac{1}{2}\right)\).

e) The graph looks like the hyperbola \(y = \frac{x-1}{x-2}\) (which can be rewritten as \(y = 1 + \frac{1}{x-2}\), a shifted version of \(\frac{1}{x}\)), with a hole at the point \((3, 2)\). The graph approaches \(y = 1\) from above for large positive \(x\) and from below for large negative \(x\), with a vertical asymptote at \(x = 2\).

Problem 7: Transformations and Inverse (xx)

Given \(f(x) = 2 \cdot \ln(x - 1) + 3\):

  1. Describe all transformations from the base function \(\ln(x)\)
  2. Find the domain and range
  3. Find the inverse function \(f^{-1}(x)\)
  4. Verify that \(f(f^{-1}(x)) = x\)

a) Starting from \(y = \ln(x)\), the transformations are applied in this order:

  1. Horizontal shift right by 1 unit: \(\ln(x) \to \ln(x - 1)\)
  2. Vertical stretch by factor 2: \(\ln(x-1) \to 2\ln(x-1)\)
  3. Vertical shift up by 3 units: \(2\ln(x-1) \to 2\ln(x-1) + 3\)

b) The argument of \(\ln\) must be positive: \(x - 1 > 0 \implies x > 1\).

  • Domain: \((1, \infty)\)
  • Range: Since \(\ln(x-1)\) takes all real values, so does \(2\ln(x-1) + 3\). The range is \((-\infty, \infty)\).

c) To find the inverse, swap \(x\) and \(y\) and solve for \(y\):

\[x = 2\ln(y - 1) + 3\]

\[x - 3 = 2\ln(y - 1)\]

\[\frac{x - 3}{2} = \ln(y - 1)\]

Exponentiate both sides:

\[y - 1 = e^{(x-3)/2}\]

\[y = e^{(x-3)/2} + 1\]

Therefore: \(f^{-1}(x) = e^{(x-3)/2} + 1\)

d) Compute \(f(f^{-1}(x))\):

\[f\!\left(e^{(x-3)/2} + 1\right) = 2 \cdot \ln\!\left(e^{(x-3)/2} + 1 - 1\right) + 3\]

\[= 2 \cdot \ln\!\left(e^{(x-3)/2}\right) + 3\]

\[= 2 \cdot \frac{x - 3}{2} + 3 = (x - 3) + 3 = x \quad \checkmark\]

This confirms the inverse is correct.

Problem 8: Exponential Growth Word Problem (xx)

A bacteria population grows according to \(N(t) = 500 \cdot e^{0.15t}\), where \(t\) is in hours.

  1. What is the initial population?
  2. How long until the population reaches 2000?
  3. What is the doubling time?
  4. At what rate is the population growing after 3 hours?

a) The initial population is \(N(0)\):

\[N(0) = 500 \cdot e^{0.15 \cdot 0} = 500 \cdot e^0 = 500 \cdot 1 = 500\]

The initial population is 500 bacteria.

b) Set \(N(t) = 2000\) and solve for \(t\):

\[2000 = 500 \cdot e^{0.15t}\]

\[4 = e^{0.15t}\]

\[\ln(4) = 0.15t\]

\[t = \frac{\ln(4)}{0.15} = \frac{1.3863}{0.15} \approx 9.24 \text{ hours}\]

The population reaches 2000 after approximately 9.24 hours.

c) The doubling time is the time for the population to go from \(500\) to \(1000\) (or equivalently, for any value to double):

\[2 = e^{0.15t}\]

\[\ln(2) = 0.15t\]

\[t = \frac{\ln(2)}{0.15} = \frac{0.6931}{0.15} \approx 4.62 \text{ hours}\]

The doubling time is approximately 4.62 hours.

d) The rate of growth is \(N'(t)\):

\[N'(t) = 500 \cdot 0.15 \cdot e^{0.15t} = 75 \cdot e^{0.15t}\]

At \(t = 3\):

\[N'(3) = 75 \cdot e^{0.15 \cdot 3} = 75 \cdot e^{0.45} = 75 \cdot 1.5683 \approx 117.6\]

The population is growing at approximately 117.6 bacteria per hour after 3 hours.

Problem 9: Logarithmic Equations with Domain (xx)

Solve and check domain restrictions:

  1. \(\log_3(x+2) + \log_3(x) = 1\)
  2. \(\ln(2x+1) - \ln(x-2) = \ln(3)\)
  3. \(2\log_5(x) = \log_5(9) + \log_5(x-2)\)

a) Domain restriction: \(x + 2 > 0\) and \(x > 0\), so \(x > 0\).

Combine using the product rule:

\[\log_3(x(x+2)) = 1\]

\[x(x+2) = 3^1 = 3\]

\[x^2 + 2x - 3 = 0\]

\[(x+3)(x-1) = 0\]

\[x = -3 \quad \text{or} \quad x = 1\]

Check domain: \(x = -3 < 0\) is rejected. Therefore \(x = 1\).

Verification: \(\log_3(3) + \log_3(1) = 1 + 0 = 1\) ✓

b) Domain restriction: \(2x + 1 > 0\) and \(x - 2 > 0\), so \(x > 2\).

Use the quotient rule:

\[\ln\!\left(\frac{2x+1}{x-2}\right) = \ln(3)\]

\[\frac{2x+1}{x-2} = 3\]

\[2x + 1 = 3(x - 2) = 3x - 6\]

\[1 + 6 = 3x - 2x\]

\[x = 7\]

Check domain: \(7 > 2\) ✓

Verification: \(\ln(15) - \ln(5) = \ln(3)\) ✓

c) Domain restriction: \(x > 0\) and \(x - 2 > 0\), so \(x > 2\).

Rewrite using logarithm rules:

\[2\log_5(x) = \log_5(9) + \log_5(x-2)\]

\[\log_5(x^2) = \log_5(9(x-2))\]

\[x^2 = 9(x - 2)\]

\[x^2 - 9x + 18 = 0\]

\[(x - 3)(x - 6) = 0\]

\[x = 3 \quad \text{or} \quad x = 6\]

Check domain: Both \(x = 3 > 2\) and \(x = 6 > 2\) are valid.

Verification for \(x = 3\): \(2\log_5(3) = \log_5(9) = \log_5(9) + \log_5(1)\) ✓

Verification for \(x = 6\): \(2\log_5(6) = \log_5(36)\) and \(\log_5(9) + \log_5(4) = \log_5(36)\) ✓

Solutions: \(x = 3\) and \(x = 6\).

Problem 10: Market Saturation Model (xxx)

A new product’s market share is modeled by \(M(t) = \frac{60}{1 + 14 \cdot e^{-0.5t}}\) (logistic model, % share, \(t\) in months).

  1. What is the initial market share at \(t = 0\)?
  2. What is the long-term equilibrium market share?
  3. When does the market share reach 30%?
  4. Show that \(M'(t) = \frac{420 \cdot e^{-0.5t}}{(1 + 14e^{-0.5t})^2}\) and find when the growth rate is maximized.
  5. Interpret the inflection point in business terms.

a) Substitute \(t = 0\):

\[M(0) = \frac{60}{1 + 14 \cdot e^0} = \frac{60}{1 + 14} = \frac{60}{15} = 4\]

The initial market share is 4%.

b) As \(t \to \infty\), \(e^{-0.5t} \to 0\):

\[\lim_{t \to \infty} M(t) = \frac{60}{1 + 14 \cdot 0} = \frac{60}{1} = 60\]

The long-term equilibrium market share is 60%. The product will never capture more than 60% of the market.

c) Set \(M(t) = 30\):

\[30 = \frac{60}{1 + 14e^{-0.5t}}\]

\[1 + 14e^{-0.5t} = \frac{60}{30} = 2\]

\[14e^{-0.5t} = 1\]

\[e^{-0.5t} = \frac{1}{14}\]

\[-0.5t = \ln\!\left(\frac{1}{14}\right) = -\ln(14)\]

\[t = \frac{2\ln(14)}{1} = 2\ln(14) \approx 2 \times 2.6391 \approx 5.28 \text{ months}\]

The market share reaches 30% after approximately 5.28 months.

d) We compute \(M'(t)\) using the quotient rule. Write \(M(t) = 60 \cdot (1 + 14e^{-0.5t})^{-1}\):

\[M'(t) = 60 \cdot (-1) \cdot (1 + 14e^{-0.5t})^{-2} \cdot 14 \cdot (-0.5) \cdot e^{-0.5t}\]

\[= 60 \cdot \frac{7 \cdot e^{-0.5t}}{(1 + 14e^{-0.5t})^2} = \frac{420 \cdot e^{-0.5t}}{(1 + 14e^{-0.5t})^2} \quad \checkmark\]

For logistic functions of the form \(\frac{L}{1 + Ae^{-kt}}\), the growth rate is maximized at the inflection point, which occurs when \(M(t) = \frac{L}{2}\):

\[M(t) = \frac{60}{2} = 30\]

From part (c), this happens at \(t = 2\ln(14) \approx 5.28\) months.

e) The inflection point at \(t \approx 5.28\) months (when market share is 30%) represents the moment when the product’s market growth transitions from accelerating to decelerating. Before this point, each month brings a larger gain in market share than the previous month (word-of-mouth, network effects are building). After this point, growth slows as the market approaches saturation. This is a critical turning point for business strategy: marketing spending may become less efficient after this point, and the company should begin focusing on retention rather than acquisition.

Problem 11: Rational Function with Hole and Asymptotes (xxx)

For \(f(x) = \frac{x^3 - x^2 - 6x}{x^2 - 4}\):

  1. Factor completely
  2. Identify holes, vertical asymptotes, and zeros
  3. Determine if there is a horizontal or oblique asymptote using polynomial long division
  4. Find the y-intercept (if it exists)
  5. Sketch the complete graph including the oblique asymptote

a) Factor both numerator and denominator:

Numerator: \(x^3 - x^2 - 6x = x(x^2 - x - 6) = x(x - 3)(x + 2)\)

Denominator: \(x^2 - 4 = (x - 2)(x + 2)\)

\[f(x) = \frac{x(x-3)(x+2)}{(x-2)(x+2)}\]

b) The common factor \((x + 2)\) cancels:

\[f(x) = \frac{x(x-3)}{x-2}, \quad x \neq -2\]

  • Hole at \(x = -2\): The value of the simplified form is \(\frac{(-2)(-2-3)}{-2-2} = \frac{(-2)(-5)}{-4} = \frac{10}{-4} = -\frac{5}{2}\). The hole is at \(\left(-2, -\frac{5}{2}\right)\).
  • Vertical asymptote at \(x = 2\) (remaining zero of denominator).
  • Zeros at \(x = 0\) and \(x = 3\) (zeros of the simplified numerator, both in the domain).

c) The simplified form is \(\frac{x^2 - 3x}{x - 2}\). Since the degree of the numerator (2) exceeds the degree of the denominator (1) by exactly 1, there is an oblique asymptote.

Perform polynomial long division:

\[x^2 - 3x \div (x - 2):\]

\[x^2 - 3x = (x - 1)(x - 2) + (-2)\]

Let us verify: \((x-1)(x-2) = x^2 - 3x + 2\), so \(x^2 - 3x = (x-1)(x-2) - 2\).

\[f(x) = x - 1 + \frac{-2}{x - 2}\]

The oblique asymptote is \(y = x - 1\).

d) \(f(0) = \frac{0 \cdot (0-3)}{0-2} = \frac{0}{-2} = 0\)

The y-intercept is \((0, 0)\), which coincides with one of the zeros.

e) Key features for the sketch:

  • Zeros at \(x = 0\) and \(x = 3\)
  • Hole at \((-2, -2.5)\) (open circle)
  • Vertical asymptote at \(x = 2\) (the function goes to \(+\infty\) as \(x \to 2^+\) and \(-\infty\) as \(x \to 2^-\))
  • Oblique asymptote \(y = x - 1\) (the graph approaches this line as \(x \to \pm\infty\))
  • The graph crosses the oblique asymptote where \(\frac{-2}{x-2} = 0\), which never happens, so the graph never crosses the oblique asymptote

Problem 12: Function Composition and Domain (xxx)

Let \(f(x) = \sqrt{x+4}\) and \(g(x) = \frac{1}{x-1}\).

  1. Find \((f \circ g)(x)\) and determine its domain
  2. Find \((g \circ f)(x)\) and determine its domain
  3. Find \(x\) such that \((f \circ g)(x) = 2\)
  4. Is \((f \circ g)(x) = (g \circ f)(x)\) for all \(x\) in the intersection of their domains? Prove or disprove.

a) \((f \circ g)(x) = f(g(x)) = \sqrt{g(x) + 4} = \sqrt{\frac{1}{x-1} + 4}\)

Simplify: \(\frac{1}{x-1} + 4 = \frac{1 + 4(x-1)}{x-1} = \frac{4x - 3}{x-1}\)

\[(f \circ g)(x) = \sqrt{\frac{4x-3}{x-1}}\]

Domain conditions:

  1. \(g(x)\) must be defined: \(x \neq 1\)
  2. The expression under the square root must be non-negative: \(\frac{4x-3}{x-1} \geq 0\)

For condition 2, analyze the sign of \(\frac{4x-3}{x-1}\). The critical points are \(x = \frac{3}{4}\) and \(x = 1\).

Interval \(4x - 3\) \(x - 1\) Fraction
\(x < \frac{3}{4}\) \(-\) \(-\) \(+\)
\(\frac{3}{4} < x < 1\) \(+\) \(-\) \(-\)
\(x > 1\) \(+\) \(+\) \(+\)

The fraction equals zero at \(x = \frac{3}{4}\).

Domain: \(\left(-\infty, \frac{3}{4}\right] \cup (1, \infty)\)

b) \((g \circ f)(x) = g(f(x)) = \frac{1}{f(x) - 1} = \frac{1}{\sqrt{x+4} - 1}\)

Domain conditions:

  1. \(f(x)\) must be defined: \(x + 4 \geq 0 \implies x \geq -4\)
  2. \(g\) must be defined at \(f(x)\): \(\sqrt{x+4} \neq 1 \implies x + 4 \neq 1 \implies x \neq -3\)

Domain: \([-4, -3) \cup (-3, \infty)\)

c) Set \((f \circ g)(x) = 2\):

\[\sqrt{\frac{4x-3}{x-1}} = 2\]

Square both sides:

\[\frac{4x-3}{x-1} = 4\]

\[4x - 3 = 4(x - 1) = 4x - 4\]

\[-3 = -4\]

This is a contradiction! There is no solution. No value of \(x\) satisfies \((f \circ g)(x) = 2\).

d) We disprove this by finding a single counterexample. Take \(x = 5\):

\[(f \circ g)(5) = \sqrt{\frac{4(5)-3}{5-1}} = \sqrt{\frac{17}{4}} = \frac{\sqrt{17}}{2} \approx 2.06\]

\[(g \circ f)(5) = \frac{1}{\sqrt{5+4} - 1} = \frac{1}{\sqrt{9} - 1} = \frac{1}{3 - 1} = \frac{1}{2}\]

Since \(\frac{\sqrt{17}}{2} \neq \frac{1}{2}\), we have shown that \((f \circ g)(x) \neq (g \circ f)(x)\) in general.

This illustrates the important principle that function composition is not commutative.

Problem 13: Oblique Asymptote Analysis (xxx)

For \(f(x) = \frac{2x^2 + 3x - 5}{x + 1}\):

  1. Perform polynomial long division to find the oblique asymptote
  2. Find all zeros of \(f\)
  3. Determine the vertical asymptote and behavior near it
  4. Find where \(f(x)\) intersects its oblique asymptote
  5. Sketch the graph showing the oblique asymptote clearly

a) Perform long division of \(2x^2 + 3x - 5\) by \(x + 1\):

\[2x^2 + 3x - 5 = (2x + 1)(x + 1) - 6\]

Check: \((2x+1)(x+1) = 2x^2 + 2x + x + 1 = 2x^2 + 3x + 1\), so \(2x^2 + 3x - 5 = (2x+1)(x+1) - 6\). ✓

Therefore:

\[f(x) = 2x + 1 - \frac{6}{x + 1}\]

The oblique asymptote is \(y = 2x + 1\).

b) Set the numerator to zero:

\[2x^2 + 3x - 5 = 0\]

Using the quadratic formula:

\[x = \frac{-3 \pm \sqrt{9 + 40}}{4} = \frac{-3 \pm \sqrt{49}}{4} = \frac{-3 \pm 7}{4}\]

\[x = \frac{4}{4} = 1 \quad \text{or} \quad x = \frac{-10}{4} = -\frac{5}{2}\]

Zeros: \(x = 1\) and \(x = -\frac{5}{2}\)

c) The vertical asymptote is at \(x = -1\) (where the denominator is zero and no cancellation occurs).

Behavior near \(x = -1\):

  • As \(x \to -1^+\): the term \(\frac{-6}{x+1} \to -\infty\), so \(f(x) \to -\infty\)
  • As \(x \to -1^-\): the term \(\frac{-6}{x+1} \to +\infty\), so \(f(x) \to +\infty\)

d) The function intersects the oblique asymptote where \(f(x) = 2x + 1\):

\[2x + 1 - \frac{6}{x+1} = 2x + 1\]

\[-\frac{6}{x+1} = 0\]

This equation has no solution, since \(-6 \neq 0\). The graph never intersects the oblique asymptote.

e) Key features for the sketch:

  • Zeros at \(x = 1\) and \(x = -2.5\)
  • Vertical asymptote at \(x = -1\)
  • Oblique asymptote \(y = 2x + 1\) (dashed line)
  • For \(x > -1\): the graph is below the oblique asymptote (since \(\frac{-6}{x+1} < 0\)), crossing the x-axis at \(x = 1\)
  • For \(x < -1\): the graph is above the oblique asymptote (since \(\frac{-6}{x+1} > 0\)), crossing the x-axis at \(x = -2.5\)
  • y-intercept: \(f(0) = \frac{-5}{1} = -5\)

Problem 14: Competing Exponential Models (xxxx)

Two competing technologies have adoption rates:

  • Technology A: \(A(t) = 100(1 - e^{-0.4t})\) (%)
  • Technology B: \(B(t) = 100(1 - e^{-0.15t})\) (%)

where \(t\) is in years.

  1. Which technology is adopted faster initially? Justify using derivatives at \(t=0\).
  2. Find the exact time when both technologies have equal adoption rates.
  3. A company invests if a technology reaches 80% adoption. How much sooner does Technology A reach this threshold compared to Technology B?
  4. A “combined market” function is \(C(t) = A(t) + B(t) - A(t) \cdot B(t)/100\). Find \(\lim_{t \to \infty} C(t)\) and interpret.

a) Compute the derivatives:

\[A'(t) = 100 \cdot 0.4 \cdot e^{-0.4t} = 40e^{-0.4t}\]

\[B'(t) = 100 \cdot 0.15 \cdot e^{-0.15t} = 15e^{-0.15t}\]

At \(t = 0\):

\[A'(0) = 40 \cdot 1 = 40 \quad \text{and} \quad B'(0) = 15 \cdot 1 = 15\]

Since \(A'(0) = 40 > 15 = B'(0)\), Technology A is adopted faster initially — its adoption rate at launch is more than 2.5 times that of Technology B.

b) Set \(A(t) = B(t)\):

\[100(1 - e^{-0.4t}) = 100(1 - e^{-0.15t})\]

\[1 - e^{-0.4t} = 1 - e^{-0.15t}\]

\[e^{-0.4t} = e^{-0.15t}\]

\[-0.4t = -0.15t\]

\[-0.25t = 0\]

\[t = 0\]

The adoption rates are equal only at \(t = 0\) (both start at 0%). For all \(t > 0\), Technology A has a higher adoption rate than Technology B since \(0.4 > 0.15\) means \(e^{-0.4t} < e^{-0.15t}\), which gives \(A(t) > B(t)\).

c) For Technology A at 80%:

\[80 = 100(1 - e^{-0.4t})\]

\[0.8 = 1 - e^{-0.4t}\]

\[e^{-0.4t} = 0.2\]

\[t_A = \frac{-\ln(0.2)}{0.4} = \frac{\ln(5)}{0.4} = \frac{1.6094}{0.4} \approx 4.02 \text{ years}\]

For Technology B at 80%:

\[e^{-0.15t} = 0.2\]

\[t_B = \frac{\ln(5)}{0.15} = \frac{1.6094}{0.15} \approx 10.73 \text{ years}\]

The difference is:

\[t_B - t_A = \ln(5)\left(\frac{1}{0.15} - \frac{1}{0.4}\right) = \ln(5) \cdot \frac{0.4 - 0.15}{0.15 \cdot 0.4} = \ln(5) \cdot \frac{0.25}{0.06} \approx 6.71 \text{ years}\]

Technology A reaches 80% approximately 6.71 years sooner than Technology B.

d) As \(t \to \infty\), \(A(t) \to 100\) and \(B(t) \to 100\):

\[\lim_{t \to \infty} C(t) = 100 + 100 - \frac{100 \cdot 100}{100} = 100 + 100 - 100 = 100\]

Interpretation: The combined market function \(C(t)\) represents the percentage of the market that has adopted at least one of the two technologies (using the inclusion-exclusion principle, treating adoption as independent events). In the long run, 100% of the market will have adopted at least one technology. The subtraction term \(\frac{A(t) \cdot B(t)}{100}\) avoids double-counting the portion of the market that adopts both.

Problem 15: Multi-Step Logarithmic Modeling (xxxx)

A company’s revenue follows \(R(t) = a \cdot \ln(bt + 1) + c\), where \(t\) is years since founding. Given: \(R(0) = 2\) million, \(R(3) = 8\) million, \(R(10) = 12\) million.

  1. Set up a system of three equations to find \(a\), \(b\), and \(c\).
  2. Solve the system (show that \(c = 2\) immediately, then solve for \(a\) and \(b\)).
  3. Predict \(R(20)\).
  4. When will revenue reach 15 million?
  5. Find \(\lim_{t \to \infty} R(t)\) and discuss whether this model is realistic for long-term prediction.

a) Substituting the three data points:

\[R(0) = a \cdot \ln(b \cdot 0 + 1) + c = a \cdot \ln(1) + c = c = 2 \quad \text{...(i)}\]

\[R(3) = a \cdot \ln(3b + 1) + c = 8 \quad \text{...(ii)}\]

\[R(10) = a \cdot \ln(10b + 1) + c = 12 \quad \text{...(iii)}\]

b) From equation (i), we immediately get \(c = 2\).

Substituting into (ii) and (iii):

\[a \cdot \ln(3b + 1) = 6 \quad \text{...(ii')}\]

\[a \cdot \ln(10b + 1) = 10 \quad \text{...(iii')}\]

Dividing (iii’) by (ii’):

\[\frac{\ln(10b + 1)}{\ln(3b + 1)} = \frac{10}{6} = \frac{5}{3}\]

\[\ln(10b + 1) = \frac{5}{3} \ln(3b + 1)\]

\[10b + 1 = (3b + 1)^{5/3}\]

This is a transcendental equation. We solve numerically. Let \(u = 3b + 1\), then \(b = \frac{u-1}{3}\) and \(10b + 1 = \frac{10(u-1)}{3} + 1 = \frac{10u - 7}{3}\).

\[\frac{10u - 7}{3} = u^{5/3}\]

Testing \(u = 4\) (i.e., \(b = 1\)): LHS \(= \frac{33}{3} = 11\), RHS \(= 4^{5/3} = (2^2)^{5/3} = 2^{10/3} \approx 10.08\). Close but not exact.

Testing \(u = 4.2\) (i.e., \(b \approx 1.067\)): LHS \(= \frac{35}{3} \approx 11.67\), RHS \(= 4.2^{5/3} \approx 10.88\). LHS > RHS.

Testing \(u = 3.5\) (i.e., \(b \approx 0.833\)): LHS \(= \frac{28}{3} \approx 9.33\), RHS \(= 3.5^{5/3} \approx 7.65\). LHS > RHS.

Testing \(u = 5\) (i.e., \(b = 4/3\)): LHS \(= \frac{43}{3} \approx 14.33\), RHS \(= 5^{5/3} \approx 14.62\). Very close, LHS \(\approx\) RHS.

Refining around \(b \approx 1.3\): \(u = 4.9\), LHS \(= \frac{42}{3} = 14\), RHS \(= 4.9^{5/3} \approx 14.0\). This gives \(b \approx 1.3\).

Using \(b \approx 1.3\), from equation (ii’):

\[a = \frac{6}{\ln(3(1.3) + 1)} = \frac{6}{\ln(4.9)} = \frac{6}{1.589} \approx 3.78\]

So the model is approximately: \(R(t) \approx 3.78 \cdot \ln(1.3t + 1) + 2\)

c) Predict \(R(20)\):

\[R(20) \approx 3.78 \cdot \ln(1.3 \cdot 20 + 1) + 2 = 3.78 \cdot \ln(27) + 2\]

\[= 3.78 \cdot 3.296 + 2 \approx 12.46 + 2 = 14.46 \text{ million}\]

The predicted revenue at year 20 is approximately 14.46 million.

d) Set \(R(t) = 15\):

\[15 = 3.78 \cdot \ln(1.3t + 1) + 2\]

\[13 = 3.78 \cdot \ln(1.3t + 1)\]

\[\ln(1.3t + 1) = \frac{13}{3.78} \approx 3.439\]

\[1.3t + 1 = e^{3.439} \approx 31.16\]

\[t = \frac{31.16 - 1}{1.3} = \frac{30.16}{1.3} \approx 23.2 \text{ years}\]

Revenue will reach 15 million after approximately 23.2 years.

e) As \(t \to \infty\):

\[\lim_{t \to \infty} R(t) = \lim_{t \to \infty} \left[a \cdot \ln(bt + 1) + c\right] = +\infty\]

Since \(\ln(bt + 1) \to \infty\) as \(t \to \infty\), the revenue grows without bound.

Discussion: While the logarithmic model fits the data well for the given time range, it is not realistic for long-term prediction because:

  1. It predicts revenue will grow indefinitely, which is unrealistic for any real company.
  2. However, growth is extremely slow for large \(t\) (logarithmic growth), which is somewhat more realistic than polynomial or exponential models.
  3. A logistic or bounded growth model would be more appropriate for long-term forecasting, as real markets have finite size.

The model is useful for medium-term projections but should be replaced with a saturation model for very long horizons.

Problem 16: Comprehensive Function Analysis (xxxx)

For \(f(x) = \frac{(x-1)(x+2)}{(x-3)} \cdot e^{-x/2}\):

  1. Find the domain
  2. Find all zeros
  3. Determine the behavior as \(x \to 3\) (from both sides)
  4. Find \(\lim_{x \to \infty} f(x)\) and \(\lim_{x \to -\infty} f(x)\)
  5. Find \(f'(x)\) and determine all critical points (set up the equation — exact solution not required, but describe the method)
  6. Discuss what makes this function’s graph particularly interesting compared to a standard rational function

a) The exponential factor \(e^{-x/2}\) is defined for all \(x \in \mathbb{R}\). The rational part requires \(x - 3 \neq 0\).

Domain: \(\mathbb{R} \setminus \{3\} = (-\infty, 3) \cup (3, \infty)\)

b) \(f(x) = 0\) when the numerator is zero and the denominator is not:

\[(x-1)(x+2) \cdot e^{-x/2} = 0\]

Since \(e^{-x/2} > 0\) for all \(x\), we need \((x-1)(x+2) = 0\):

\[x = 1 \quad \text{or} \quad x = -2\]

Both are in the domain. Zeros: \(x = 1\) and \(x = -2\).

c) Write \(f(x) = \frac{(x-1)(x+2)}{x-3} \cdot e^{-x/2}\).

At \(x = 3\), the numerator is \((3-1)(3+2) \cdot e^{-3/2} = 10e^{-3/2} \approx 2.23 > 0\), a fixed positive number.

  • As \(x \to 3^+\): \((x - 3) \to 0^+\), so \(\frac{10e^{-3/2}}{x-3} \to +\infty\). Therefore \(f(x) \to +\infty\).
  • As \(x \to 3^-\): \((x - 3) \to 0^-\), so \(\frac{10e^{-3/2}}{x-3} \to -\infty\). Therefore \(f(x) \to -\infty\).

There is a vertical asymptote at \(x = 3\).

d) For \(x \to +\infty\):

The rational part \(\frac{(x-1)(x+2)}{x-3} = \frac{x^2 + x - 2}{x - 3}\) grows like \(x\) for large \(x\) (degree 2 over degree 1). Meanwhile, \(e^{-x/2} \to 0\) exponentially fast. The exponential decay dominates any polynomial growth:

\[\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{x^2 + x - 2}{x - 3} \cdot e^{-x/2} = 0\]

For \(x \to -\infty\):

The rational part \(\frac{x^2 + x - 2}{x - 3}\) behaves like \(\frac{x^2}{x} = x \to -\infty\). Meanwhile, \(e^{-x/2} = e^{|x|/2} \to +\infty\).

\[\lim_{x \to -\infty} f(x) = (-\infty) \cdot (+\infty) = -\infty\]

More precisely, \(|f(x)|\) grows faster than any polynomial as \(x \to -\infty\), so \(f(x) \to -\infty\).

e) Let us write \(f(x) = R(x) \cdot e^{-x/2}\) where \(R(x) = \frac{(x-1)(x+2)}{x-3}\).

Using the product rule:

\[f'(x) = R'(x) \cdot e^{-x/2} + R(x) \cdot \left(-\frac{1}{2}\right) e^{-x/2} = e^{-x/2}\left[R'(x) - \frac{1}{2}R(x)\right]\]

First, find \(R'(x)\) using the quotient rule. Let \(N(x) = (x-1)(x+2) = x^2 + x - 2\) and \(D(x) = x - 3\):

\[R'(x) = \frac{N'(x) \cdot D(x) - N(x) \cdot D'(x)}{[D(x)]^2} = \frac{(2x+1)(x-3) - (x^2+x-2)(1)}{(x-3)^2}\]

Expanding the numerator:

\[(2x+1)(x-3) - (x^2+x-2) = 2x^2 - 6x + x - 3 - x^2 - x + 2 = x^2 - 6x - 1\]

\[R'(x) = \frac{x^2 - 6x - 1}{(x-3)^2}\]

Now set \(f'(x) = 0\). Since \(e^{-x/2} > 0\) always, we need:

\[R'(x) - \frac{1}{2}R(x) = 0\]

\[\frac{x^2 - 6x - 1}{(x-3)^2} = \frac{1}{2} \cdot \frac{x^2 + x - 2}{x - 3}\]

Multiply both sides by \((x-3)^2\):

\[x^2 - 6x - 1 = \frac{1}{2}(x^2 + x - 2)(x - 3)\]

\[2(x^2 - 6x - 1) = (x^2 + x - 2)(x - 3)\]

\[2x^2 - 12x - 2 = x^3 - 3x^2 + x^2 - 3x - 2x + 6 = x^3 - 2x^2 - 5x + 6\]

\[0 = x^3 - 4x^2 + 7x + 8\]

This is a cubic equation. Finding its roots requires numerical methods (e.g., Newton’s method) or a graphing calculator. The method would be:

  1. Test rational roots \(\pm 1, \pm 2, \pm 4, \pm 8\): none are roots (check: \(p(1) = 12\), \(p(-1) = 0\) — let us verify: \((-1)^3 - 4(-1)^2 + 7(-1) + 8 = -1 - 4 - 7 + 8 = -4 \neq 0\)). So no rational roots.
  2. Use numerical methods to find approximately where the cubic changes sign, yielding the critical points.

f) This function is particularly interesting because:

  1. Asymmetric end behavior: Unlike a standard rational function (which behaves the same in both directions, or symmetrically), this function approaches 0 as \(x \to +\infty\) but diverges to \(-\infty\) as \(x \to -\infty\). This is due to the exponential factor \(e^{-x/2}\).

  2. Damped oscillation-like behavior on the right: For large positive \(x\), the rational part grows but the exponential decay forces the function toward zero, creating a “damped” effect where the function may have local extrema before settling to zero.

  3. Exponential amplification on the left: For large negative \(x\), the exponential factor \(e^{-x/2}\) grows without bound, amplifying the rational function’s behavior and causing rapid divergence.

  4. The vertical asymptote at \(x = 3\) is modulated: Unlike a pure rational function where the asymptotic behavior is symmetric about the vertical asymptote, here the exponential factor makes the function approach infinity at different rates on either side.

  5. No horizontal asymptote on the left: Standard rational functions always have horizontal or oblique asymptotes. The exponential factor destroys any such asymptote for \(x \to -\infty\).

This blend of rational and exponential behavior creates a function that is far richer and more complex than either component alone.