Tasks 09-07 - Final Recap: Function Analysis & Probability

Section 09: Exam Preparation

Work on these problems as in the exam: write down every step. The points show what each part is worth.

FSP Problem 1: A Cubic Function and a Line [24 pts. total]

Consider the functions

\[f(x) = \frac{1}{2}x^3 - 9x^2 + 48x - 72 \qquad \text{and} \qquad g(x) = -\frac{3}{2}x + 9\]

Cubic f and the falling dashed line g on a half-unit grid. f rises steeply to a high point at (4, 8), falls through (6, 0) to a low point at (8, −8) and rises steeply again. The line g meets f three times, at x = 3, 6 and 9, and the two graphs enclose two regions.
Figure 1: Graphs of f(x) and g(x)

Part a)

Verify computationally that \(f(x)\) possesses a maximum at \(x = 4\).

[4 pts.]

Step 1: \(f'(4) = 0\)?

\[f'(x) = \frac{3}{2}x^2 - 18x + 48 = \frac{3}{2}(x^2 - 12x + 32) = \frac{3}{2}(x - 4)(x - 8)\]

\[f'(4) = \frac{3}{2} \cdot 0 \cdot (-4) = 0 \quad \checkmark\]

Step 2: \(f''(4) < 0\)?

\[f''(x) = 3x - 18 \qquad f''(4) = 12 - 18 = -6 < 0 \quad \checkmark\]

Step 3: maximum value: \(f(4) = 32 - 144 + 192 - 72 = 8\)

\[\boxed{\text{Local maximum at } (4, 8)}\]

Grading: 1 Pt. \(f'(x)\), 1 Pt. \(f'(4) = 0\), 1 Pt. \(f''(4) = -6 < 0\), 1 Pt. maximum \((4, 8)\).

Part b)

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

[3 pts.]

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

\[f_1(x) = f(-x) = \frac{1}{2}(-x)^3 - 9(-x)^2 + 48(-x) - 72 = -\frac{1}{2}x^3 - 9x^2 - 48x - 72\]

Reason: every point \((a, f(a))\) maps to \((-a, f(a))\), so the graph is mirrored horizontally about the \(y\)-axis. Odd powers of \(x\) change their sign, even powers and the constant stay.

\[\boxed{f_1(x) = -\frac{1}{2}x^3 - 9x^2 - 48x - 72}\]

Grading: 1 Pt. \(f_1(x) = f(-x)\), 1 Pt. simplified term, 1 Pt. reason.

Part c)

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

Hint: The figure suggests three intersection points. Check one of them by substitution and divide by the matching linear factor (or use the polynomial solver, MENU → A → 2, degree 3).

[9 pts.]

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

\[\frac{1}{2}x^3 - 9x^2 + 48x - 72 = -\frac{3}{2}x + 9 \iff \frac{1}{2}x^3 - 9x^2 + \frac{99}{2}x - 81 = 0 \iff x^3 - 18x^2 + 99x - 162 = 0\]

The figure suggests \(x = 3\): \(27 - 162 + 297 - 162 = 0 \; \checkmark\)

Polynomial division: \((x^3 - 18x^2 + 99x - 162) : (x - 3) = x^2 - 15x + 54 = (x - 6)(x - 9)\)

\[x = 3, \quad x = 6, \quad x = 9\]

Difference function: \(h(x) = f(x) - g(x) = \frac{1}{2}(x - 3)(x - 6)(x - 9) = \frac{1}{2}x^3 - 9x^2 + \frac{99}{2}x - 81\)

Which graph lies above?

  • \(3 < x < 6\) (test \(x = 4\)): \(h(4) = \frac{1}{2} \cdot 1 \cdot (-2) \cdot (-5) = 5 > 0\) (\(f\) above \(g\))
  • \(6 < x < 9\) (test \(x = 7\)): \(h(7) = \frac{1}{2} \cdot 4 \cdot 1 \cdot (-2) = -4 < 0\) (\(g\) above \(f\))

Antiderivative of \(h\): \(H(x) = \frac{x^4}{8} - 3x^3 + \frac{99}{4}x^2 - 81x\) with

\[H(3) = 10.125 - 81 + 222.75 - 243 = -91.125, \quad H(6) = 162 - 648 + 891 - 486 = -81, \quad H(9) = 820.125 - 2187 + 2004.75 - 729 = -91.125\]

Region 1 (\(3 \leq x \leq 6\)):

\[A_1 = \int_3^6 h(x) \, dx = H(6) - H(3) = -81 - (-91.125) = 10.125\]

Region 2 (\(6 \leq x \leq 9\)):

\[A_2 = \left| \int_6^9 h(x) \, dx \right| = \left| H(9) - H(6) \right| = \left| -91.125 - (-81) \right| = 10.125\]

\[\boxed{A_1 = A_2 = 10.125 \quad \checkmark}\]

Grading: 1 Pt. equation \(f(x) = g(x)\), 1 Pt. \(x = 3\) checked, 1 Pt. polynomial division, 1 Pt. \(x = 6\) and \(x = 9\), 1 Pt. which graph lies above, 1 Pt. antiderivative \(H\), 2 Pts. \(A_1 = 10.125\) and \(A_2 = 10.125\) with the absolute value, 1 Pt. conclusion.

Part d)

The following figure shows the graph of a function \(h(x)\) of degree 3. Its local minimum and its local maximum lie on grid points.

Curve of h on a half-unit grid from x = −1 to 8.5: it falls from about (−1, 2) to a local minimum at (1, −1), rises to a local maximum at (6, 4) and falls again to about −1 at x = 8.5.
Figure 2: Graph of h(x)

Argue whether the following statements are true or false.

(1) The graph of \(h\) is concave down on the interval \(4 < x < 6\).

(2) \(h''(1) = 0\), because the tangent of \(h\) at \(x = 1\) is horizontal.

(3) The graph of \(h'\) is a parabola that opens downward and has its highest point at \(x = 3.5\).

(4) The tangent of \(h\) at \(x = 3.5\) is less steep than the secant through the local minimum and the local maximum of \(h\).

[8 pts. total, 2 pts. each]

From the graph: local minimum \((1, -1)\), local maximum \((6, 4)\). So \(h'(1) = h'(6) = 0\). \(h\) has degree 3, so \(h'\) is a parabola with the zeros 1 and 6. Its vertex lies halfway between the zeros, at \(x = 3.5\) (Session 03-03). There \(h'\) has its extremum, so \(h\) has its inflection point \((3.5, 1.5)\). The graph is concave up left of \(x = 3.5\) and concave down right of it.

(1) True. \(h'\) is a parabola with its highest point at \(x = 3.5\) (see (3)), so \(h'\) decreases on \(3.5 < x < 6\). Hence \(h'' < 0\) on \(4 < x < 6\), and the graph is concave down there.

(2) False. A horizontal tangent means \(h'(1) = 0\), not \(h''(1) = 0\). At the minimum the graph is concave up, so \(h''(1) > 0\).

(3) True. \(h\) has degree 3, so \(h'\) has degree 2: its graph is a parabola. \(h'\) is negative left of \(x = 1\), positive between 1 and 6 and negative right of 6, so the parabola opens downward with the zeros 1 and 6. Its highest point lies halfway between them, at \(x = 3.5\): there \(h\) rises most steeply, at its inflection point.

(4) False. The secant through \((1, -1)\) and \((6, 4)\) has the slope \(\frac{4 - (-1)}{6 - 1} = 1\). At the inflection point \(x = 3.5\), \(h\) rises more steeply than anywhere else, so its tangent there is steeper than the secant. From the graph: the tangent at \((3.5, 1.5)\) rises by about 1.5 per unit.

For checking: the figure shows \(h(x) = \frac{1}{25}\left(-2x^3 + 21x^2 - 36x - 8\right)\) with \(h'(x) = -\frac{6}{25}(x - 1)(x - 6)\), \(h''(x) = \frac{1}{25}(-12x + 42)\), \(h''(1) = 1.2 > 0\) and \(h'(3.5) = 1.5 > 1\).

Grading: per statement 1 Pt. true or false, 1 Pt. reason.

FSP Problem 2: An Exponential Function with Two Parameters [17 pts. total]

The function shown in Figure 3 belongs to the function \(j(x) = (x + p) \cdot e^x + q\) with real parameters \(p\) and \(q\). The graph passes through two grid points.

Curve of j on a half-unit grid from x = −4: it starts nearly flat at about 0.9, falls through the grid point (0, −1) to a low point of about −1.7 near x = 1 and then rises steeply through the grid point (2, 1).
Figure 3: Graph of j(x) = (x + p) eˣ + q

Part a)

Determine \(p\) and \(q\) computationally for the function shown in Figure 3.

[5 pts.]

Reading from the graph: the graph passes through the grid points \((0, -1)\) (the \(y\)-intercept) and \((2, 1)\).

Setting up the system:

\[j(0) = (0 + p) \cdot e^0 + q = p + q = -1 \quad (I)\]

\[j(2) = (2 + p) \cdot e^2 + q = 1 \quad (II)\]

\((II) - (I)\) removes \(q\):

\[(2 + p)e^2 - p = 2 \iff p\left(e^2 - 1\right) = 2 - 2e^2 = -2\left(e^2 - 1\right) \iff p = -2\]

From \((I)\): \(q = -1 - p = 1\).

\[\boxed{p = -2, \quad q = 1, \qquad j(x) = (x - 2) \cdot e^x + 1}\]

Check: for \(x \to -\infty\) the graph approaches the line \(y = 1 = q\) ✓.

Grading: 1 Pt. two points read from the graph, 1 Pt. equation \((I)\), 1 Pt. equation \((II)\), 1 Pt. \(p = -2\), 1 Pt. \(q = 1\).

Part b)

(for verification: \(p = -2\) and \(q = 1\), so \(j(x) = (x - 2) \cdot e^x + 1\). Use this function from now on.)

Investigate the asymptotic behavior of \(j\).

[3 pts.]

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 \((x - 2) e^x \to 0\):

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

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

Grading: 1 Pt. \(j \to +\infty\) for \(x \to +\infty\), 1 Pt. \((x - 2)e^x \to 0\) for \(x \to -\infty\), 1 Pt. horizontal asymptote \(y = 1\).

Part c)

Determine the inflection point of \(j\).

[5 pts.]

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, \(x = 0\).

Sufficient condition: \(j'''(x) = e^x \cdot x + e^x = e^x(x + 1)\), so \(j'''(0) = 1 \neq 0\). Equivalently, \(j''(x) = e^x \cdot x\) changes its sign from \(-\) to \(+\) at \(x = 0\).

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

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

The inflection point is the \(y\)-intercept that you read in a).

Grading: 1 Pt. \(j'(x)\), 1 Pt. \(j''(x)\), 1 Pt. \(x = 0\) with \(e^x > 0\), 1 Pt. sufficient condition, 1 Pt. point \((0, -1)\).

Part d)

Find the antiderivative of \(j(x)\) and show your work.

[4 pts.]

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:

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

Check: \(J'(x) = e^x + (x - 3)e^x + 1 = (x - 2)e^x + 1 = j(x)\) ✓

Grading: 1 Pt. split into two integrals, 1 Pt. choice of \(u\) and \(v'\), 1 Pt. \(\int (x - 2)e^x \, dx = (x - 3)e^x\), 1 Pt. \(J(x)\) with \(+ x + C\).

FSP Problem 3: A Reading Club at a Bookstore [24 pts. total]

A bookstore has a reading club for its customers. Round probabilities to four decimal places.

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.

Use the following 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

[6 pts.]

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

Grading: 1 Pt. column sums 25,000 and 75,000, 2 Pts. \(R \cap C\) and \(\bar{R} \cap C\), 2 Pts. \(R \cap \bar{C}\) and \(\bar{R} \cap \bar{C}\), 1 Pt. row sums 31,250 and 68,750.

Part b)

State the businesslike meaning of the quantities \(\bar{R}\) and \(\bar{C} \cap \bar{R}\).

[2 pts.]

  • \(\bar{R}\): the person does not read daily (whether a customer of the bookstore or not).
  • \(\bar{C} \cap \bar{R}\): the person is not a customer of this bookstore and does not read daily. For the bookstore, this is the group that the reading club hardly reaches.

Grading: 1 Pt. per quantity.

Part c)

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

[3 pts.]

Joint probability:

\[P(\bar{C} \cap \bar{R}) = P(\bar{R} \mid \bar{C}) \cdot P(\bar{C}) = 0.85 \times 0.75 = 0.6375\]

This matches the table: \(\tfrac{63{,}750}{100{,}000} = 0.6375\).

Conditional probability:

\[P(\bar{C} \mid \bar{R}) = \frac{P(\bar{C} \cap \bar{R})}{P(\bar{R})} = \frac{63{,}750}{68{,}750} = \frac{51}{55} \approx 0.9273\]

About 92.73% of the people who do not read daily are not customers of the bookstore.

Grading: 1 Pt. \(P(\bar{C} \cap \bar{R}) = 0.6375\), 1 Pt. formula of the conditional probability, 1 Pt. \(P(\bar{C} \mid \bar{R}) \approx 0.9273\).

Part d)

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

[3 pts.]

Bayes’ theorem (read directly from the table):

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

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

Only 25% of all people are customers, but 64% of the daily readers are. Reading daily is much more common among customers.

Grading: 1 Pt. \(P(C \mid R)\) as the probability asked for, 1 Pt. \(\frac{20{,}000}{31{,}250}\), 1 Pt. \(0.64\).

Part e)

For a sample of 10 people, determine the likelihood (probability) 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.

[2 pts.]

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

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

Grading: 1 Pt. binomial term with \(\binom{10}{3} = 120\), 1 Pt. \(0.2503\).

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

[4 pts.]

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

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

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

Grading: 1 Pt. sum from \(X = 0\) to \(X = 3\), 2 Pts. \(P(X = 0)\), \(P(X = 1)\) and \(P(X = 2)\), 1 Pt. \(0.7759\).

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

[4 pts.]

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

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

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

Grading: 1 Pt. sum from \(X = 2\) to \(X = 4\), 2 Pts. \(P(X = 4)\) with \(\binom{10}{4} = 210\), 1 Pt. \(0.6778\).