Session 09-06: Tasks

Final Recap - Function Analysis & Probability

Exam-Style Problem 1: Function Analysis

Consider the function

\[f(x) = -2x(x - 3)^2 + 4 \qquad \text{and} \qquad g(x) = -2x + 4\]

Figure 1: Graphs of f(x) and g(x)

Part a)

Verify computationally that \(f(x)\) possesses a maximum at \(x = 3\).

Expand: \(f(x) = -2x(x^2 - 6x + 9) + 4 = -2x^3 + 12x^2 - 18x + 4\)

Step 1: \(f'(3) = 0\)?

\[f'(x) = -6x^2 + 24x - 18 = -6(x^2 - 4x + 3) = -6(x - 1)(x - 3)\]

\[f'(3) = -6 \cdot 2 \cdot 0 = 0 \quad \checkmark\]

Step 2: \(f''(3) < 0\)?

\[f''(x) = -12x + 24 \qquad f''(3) = -36 + 24 = -12 < 0 \quad \checkmark\]

Step 3: maximum value: \(f(3) = -2 \cdot 3 \cdot 0 + 4 = 4\)

\[\boxed{\text{Local maximum at } (3, 4)}\]

Part b)

Give a function \(f_1(x)\) which is a mirror image of \(f(x)\) at the \(y\)-axis and give a reason for your choice.

A reflection across the \(y\)-axis is obtained by replacing \(x\) with \(-x\):

\[f_1(x) = f(-x) = -2(-x)(-x - 3)^2 + 4 = 2x(x + 3)^2 + 4\]

Reason: every point \((a, f(a))\) maps to \((-a, f(a))\), so the graph is mirrored horizontally about the \(y\)-axis.

\[\boxed{f_1(x) = 2x(x + 3)^2 + 4}\]

Part c)

The functions \(f(x)\) and \(g(x)\) enclose two regions. Show computationally that both regions have equal size.

Find intersections: \(f(x) = g(x)\):

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

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

\[-2x \big[ (x - 3)^2 - 1 \big] = 0\]

\[-2x(x^2 - 6x + 8) = 0 \;\Rightarrow\; -2x(x - 2)(x - 4) = 0\]

\[x = 0, \quad x = 2, \quad x = 4\]

Difference function: \(h(x) = f(x) - g(x) = -2x(x - 2)(x - 4) = -2x^3 + 12x^2 - 16x\)

Sign check:

  • \(0 < x < 2\) (test \(x = 1\)): \(h(1) = -2 \cdot 1 \cdot (-1)(-3) = -6 < 0\) (\(g\) above \(f\))
  • \(2 < x < 4\) (test \(x = 3\)): \(h(3) = -2 \cdot 3 \cdot 1 \cdot (-1) = 6 > 0\) (\(f\) above \(g\))

Region 1 (\(0 \leq x \leq 2\)):

\[A_1 = \left| \int_0^2 (-2x^3 + 12x^2 - 16x) \, dx \right| = \left| \left[ -\frac{x^4}{2} + 4x^3 - 8x^2 \right]_0^2 \right|\]

\[= \left| -8 + 32 - 32 \right| = 8\]

Region 2 (\(2 \leq x \leq 4\)):

\[A_2 = \left| \left[ -\frac{x^4}{2} + 4x^3 - 8x^2 \right]_2^4 \right| = \left| (-128 + 256 - 128) - (-8) \right| = 8\]

\[\boxed{A_1 = A_2 = 8 \quad \checkmark}\]

Exam-Style Problem 2: Exponential Function

Consider the function \(j(x) = (x - 2) \cdot e^x + 1\).

Figure 2: Graph of j(x) = (x − 2) eˣ + 1

Part a)

Determine the end behavior of \(j(x)\) at positive and negative infinity analytically.

As \(x \to +\infty\): \((x - 2) \to \infty\) and \(e^x \to \infty\), so

\[\lim_{x \to +\infty} j(x) = +\infty\]

As \(x \to -\infty\): exponential decay beats linear growth, so \((x - 2) e^x \to 0\):

\[\lim_{x \to -\infty} j(x) = 0 + 1 = 1\]

\[\boxed{j \to +\infty \text{ at } +\infty; \quad y = 1 \text{ horizontal asymptote at } -\infty}\]

Part b)

Compute the inflection point of \(j(x)\).

First and second derivative (product rule):

\[j'(x) = e^x + (x - 2) e^x = e^x (x - 1)\]

\[j''(x) = e^x (x - 1) + e^x = e^x \cdot x\]

Solve \(j''(x) = 0\): since \(e^x > 0\) always, \(x = 0\).

\(y\)-coordinate: \(j(0) = (-2) \cdot 1 + 1 = -1\).

\[\boxed{\text{Inflection point at } (0, -1)}\]

Part c)

Compute the antiderivative of \(j(x) = (x - 2) \cdot e^x + 1\).

Split the integral:

\[\int j(x) \, dx = \int (x - 2) e^x \, dx + \int 1 \, dx\]

Integration by parts on \((x - 2) e^x\) with \(u = x - 2\), \(v' = e^x\) (so \(u' = 1\), \(v = e^x\)):

\[\int (x - 2) e^x \, dx = (x - 2) e^x - \int e^x \, dx = (x - 2) e^x - e^x = (x - 3) e^x\]

Combine:

\[\boxed{J(x) = (x - 3) e^x + x + C}\]

Exam-Style Problem 3: Probability & Statistics

A bookstore has a reading club for its customers.

Consider the following probabilities:

  • The probability of a random person being a customer of this bookstore is 25%.
  • A customer of this bookstore has an 80% chance of reading daily.
  • A non-customer has a 15% chance of reading daily.

Notation:

  • \(C\): “is a customer of this bookstore”
  • \(R\): “reads daily”

Part a)

Display the fractions in a contingency table for a population of 100,000 people. Fill in the blank fields.

\(C\) \(\bar{C}\) Sum
\(R\)
\(\bar{R}\)
Sum 100,000

Given: \(P(C) = 0.25\), \(P(R \mid C) = 0.80\), \(P(R \mid \bar{C}) = 0.15\).

Out of 100,000 people:

  • \(C\): \(100{,}000 \times 0.25 = 25{,}000\)
  • \(\bar{C}\): \(75{,}000\)
  • \(R \cap C\): \(25{,}000 \times 0.80 = 20{,}000\)
  • \(\bar{R} \cap C\): \(5{,}000\)
  • \(R \cap \bar{C}\): \(75{,}000 \times 0.15 = 11{,}250\)
  • \(\bar{R} \cap \bar{C}\): \(63{,}750\)
\(C\) \(\bar{C}\) Sum
\(R\) 20,000 11,250 31,250
\(\bar{R}\) 5,000 63,750 68,750
Sum 25,000 75,000 100,000

Part b)

Compute the probabilities \(P(\bar{C} \cap R)\) and \(P(R \mid \bar{C})\).

Joint probability:

\[P(\bar{C} \cap R) = P(R \mid \bar{C}) \cdot P(\bar{C}) = 0.15 \times 0.75 = 0.1125\]

\[\boxed{P(\bar{C} \cap R) = 0.1125 = 11.25\%}\]

This matches the table: \(\tfrac{11{,}250}{100{,}000} = 0.1125\).

Conditional probability:

\[P(R \mid \bar{C}) = \frac{P(R \cap \bar{C})}{P(\bar{C})} = \frac{11{,}250}{75{,}000} = 0.15\]

\[\boxed{P(R \mid \bar{C}) = 15\%}\]

Part c)

A person reads daily. Compute the probability that this person is a customer of the bookstore.

Bayes’ theorem (read directly from the table):

\[P(C \mid R) = \frac{|C \cap R|}{|R|} = \frac{20{,}000}{31{,}250} = 0.64\]

\[\boxed{P(C \mid R) = 64\%}\]

Part d)

For a sample of 10 people determine the likelihood of the following events. This is a binomial distribution with \(n = 10\) and \(p = P(C) = 0.25\).

\[P(X = k) = \binom{10}{k} (0.25)^k (0.75)^{10 - k}\]

1. Exactly 3 persons are customers of this bookstore.

\[P(X = 3) = \binom{10}{3} (0.25)^3 (0.75)^7 = 120 \cdot 0.015625 \cdot 0.1335\]

\[\boxed{P(X = 3) \approx 0.2503 \approx 25.0\%}\]

2. Not more than 3 persons are customers of this bookstore.

\[P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)\]

  • \(P(X = 0) = (0.75)^{10} \approx 0.0563\)
  • \(P(X = 1) = 10 \cdot 0.25 \cdot (0.75)^9 \approx 0.1877\)
  • \(P(X = 2) = 45 \cdot 0.0625 \cdot (0.75)^8 \approx 0.2816\)
  • \(P(X = 3) \approx 0.2503\) (from above)

\[\boxed{P(X \leq 3) \approx 0.7759 \approx 77.6\%}\]

3. At least 2 but not more than 4 persons are customers of this bookstore.

\[P(2 \leq X \leq 4) = P(X = 2) + P(X = 3) + P(X = 4)\]

  • \(P(X = 2) \approx 0.2816\)
  • \(P(X = 3) \approx 0.2503\)
  • \(P(X = 4) = 210 \cdot (0.25)^4 \cdot (0.75)^6 \approx 0.1460\)

\[\boxed{P(2 \leq X \leq 4) \approx 0.6778 \approx 67.8\%}\]