Session 09-06 - Final Recap: Function Analysis & Probability

Section 09: Exam Preparation

Author

Dr. Nikolai Heinrichs & Dr. Tobias Vlćek

Concept Refresher - Function Analysis - 25 Minutes

Verifying Extrema (Second Derivative)

To prove that \(f\) has a maximum or minimum at \(x = a\):

Show \(f'(a) = 0\) (stationary point)
Compute \(f''(a)\):
- \(f''(a) < 0\) \(\Rightarrow\) local maximum
- \(f''(a) > 0\) \(\Rightarrow\) local minimum
- \(f''(a) = 0\) \(\Rightarrow\) inconclusive; check sign of \(f'\)
Compute \(f(a)\) to give the actual extremum value

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The exam often asks: “Verify computationally that \(f\) has a maximum at \(x = a\).” All three steps are needed for full points.

Symmetry & Reflections

Transformation	Formula	Meaning
Reflection at \(y\)-axis	\(f(-x)\)	Mirror left/right
Reflection at \(x\)-axis	\(-f(x)\)	Mirror up/down
Vertical shift	\(f(x) + c\)	Move up by \(c\)
Horizontal shift	\(f(x - c)\)	Move right by \(c\)

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Quick: If \(f(x) = 2x(x-3)^2 - 4\), then the reflection at the \(y\)-axis is?

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

Areas Between Curves

Steps:

Find intersection points: solve \(f(x) = g(x)\)
Form the difference \(h(x) = f(x) - g(x)\)
Determine which function is on top in each interval (test a value)
Integrate \(|h(x)|\) over each region; areas are always positive

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\[A = \int_{a}^{b} \big| f(x) - g(x) \big| \, dx\]

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If \(h\) changes sign, split the integral at each zero. Otherwise the positive and negative parts cancel.

Areas Between Curves: Example

Find the area enclosed between \(f(x) = x^2\) and \(g(x) = 2x\) on \([0, 2]\):

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Step 1, intersections: \(x^2 = 2x \Rightarrow x(x-2) = 0 \Rightarrow x = 0, 2\)

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Step 2, top function: at \(x = 1\): \(g(1) = 2 > 1 = f(1)\); so \(g\) is on top.

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\[A = \int_0^2 (2x - x^2) \, dx = \left[ x^2 - \frac{x^3}{3} \right]_0^2 = 4 - \frac{8}{3} = \frac{4}{3}\]

True/False Statements From Graphs

To assess claims about \(f\), \(f'\), \(f''\) from a graph, recall:

\(f'(a) > 0 \Leftrightarrow f\) is increasing at \(a\)
\(f''(a) > 0 \Leftrightarrow f\) is concave up at \(a\) (smiley)
\(f''(a) < 0 \Leftrightarrow f\) is concave down at \(a\) (frowny)
Tangent slope at \(a\) vs. secant slope \(\frac{f(b) - f(a)}{b - a}\) over \([a, b]\)

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Mean Value Theorem: somewhere in \((a, b)\) the tangent slope equals the secant slope. So at \(x = a\) the tangent can be smaller, equal, or bigger.

Exponential Function Basics

Property	\(e^x\)
Domain / Range	\(\mathbb{R}\) / \((0, \infty)\)
End behavior	\(\to 0\) as \(x \to -\infty\); \(\to \infty\) as \(x \to \infty\)

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Combined with chain & product rule:

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\[\big( e^{g(x)} \big)' = e^{g(x)} \cdot g'(x), \qquad \big( (x + p) e^x \big)' = e^x(x + p + 1)\]

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End behavior of \((x + p) e^x - q\):

\(x \to +\infty\): linear \(\times\) exp blows up; \(j \to +\infty\)
\(x \to -\infty\): exponential decay wins; \((x+p)e^x \to 0\), so \(j \to -q\)

Reading Parameters From a Graph

Strategy:

Identify readable points (intercepts, asymptote levels, special values)
Each readable point gives one equation in the unknowns
Solve the resulting system, then round if instructed

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Example with \(j(x) = (x + p) e^x - q\):

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If \(j(0) = -3\) (read from \(y\)-intercept), then \(p - q = -3\).

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One equation, two unknowns; need a second point.

Inflection Points

Recipe:

Compute \(j''(x)\)
Solve \(j''(x) = 0\)
Verify sign change of \(j''\) (or compute \(j'''\))
Compute \(y\)-coordinate \(j(x_0)\)

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For \(j(x) = (x + p) e^x - q\):

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\[j'(x) = e^x(x + p + 1), \qquad j''(x) = e^x(x + p + 2)\]

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Since \(e^x > 0\) always: \(j''(x) = 0 \Leftrightarrow x = -(p + 2)\).

Integration by Parts

\[\int u \cdot v' \, dx = u \cdot v - \int u' \cdot v \, dx\]

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Choosing: pick \(u\) so that \(u'\) is simpler; pick \(v'\) so it is easy to integrate.

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Example: \(\int (x + 1) e^x \, dx\)

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Pick \(u = x + 1\), \(v' = e^x\); then \(u' = 1\), \(v = e^x\):

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

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

Concept Refresher - Probability - 15 Minutes

Contingency Tables

For two events \(A\), \(B\) on a population of size \(N\):

	\(A\)	\(\bar{A}\)	Sum
\(B\)	\(N \cdot P(A \cap B)\)	\(N \cdot P(\bar{A} \cap B)\)	\(N \cdot P(B)\)
\(\bar{B}\)	\(N \cdot P(A \cap \bar{B})\)	\(N \cdot P(\bar{A} \cap \bar{B})\)	\(N \cdot P(\bar{B})\)
Sum	\(N \cdot P(A)\)	\(N \cdot P(\bar{A})\)	\(N\)

Conditional Probability & Bayes

Definition: \(P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}\)

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Bayes’ theorem:

\[P(A \mid B) = \frac{P(B \mid A) \cdot P(A)}{P(B)}\]

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On the exam, you can usually read \(P(A \mid B)\) directly off the contingency table: \[P(A \mid B) = \frac{\text{count in } A \cap B}{\text{count in } B \text{ row/column}}\]

Binomial Distribution

Setting: \(n\) independent trials, each with success probability \(p\).

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\[P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}\]

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Common variants:

“Exactly \(k\)”: one term
“At most \(k\)”: sum from \(0\) to \(k\)
“At least \(k\)”: \(1 - P(X \leq k - 1)\)

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Using the complement is often faster: e.g. \(P(X \geq 1) = 1 - P(X = 0)\).

Coffee Break - 15 Minutes

Walkthrough: Function Analysis - 20 Minutes

The Problem

Consider \(f(x) = 4x(x - 3)^2 - 8\) and \(g(x) = 4x - 8\).

Figure 1: f(x) = 4x(x-3)² - 8 and g(x) = 4x - 8

Verify the Maximum at \(x = 1\)

Expand: \(f(x) = 4x(x^2 - 6x + 9) - 8 = 4x^3 - 24x^2 + 36x - 8\)

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

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\[f'(x) = 12x^2 - 48x + 36 = 12(x - 1)(x - 3) \qquad f'(1) = 0 \quad \checkmark\]

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Step 2: \(f''(1) < 0\)?

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\[f''(x) = 24x - 48 \qquad f''(1) = -24 < 0 \quad \checkmark\]

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Step 3: maximum value: \(f(1) = 4 \cdot 1 \cdot 4 - 8 = 8\)

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\[\boxed{\text{Local maximum at } (1, 8)}\]

Mirror Image at the \(y\)-Axis

\[f(x) = 4x(x^2 - 6x + 9) - 8\]

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Replace \(x\) with \(-x\):

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\[f_1(x) = f(-x) = 4(-x)(-x - 3)^2 - 8 = -4x(x + 3)^2 - 8\]

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Reasoning: every point \((a, f(a))\) becomes \((-a, f(a))\), mirroring the graph horizontally about the \(y\)-axis.

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\[\boxed{f_1(x) = -4x(x + 3)^2 - 8}\]

Find the Intersections

Set \(f(x) = g(x)\):

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\[4x(x - 3)^2 - 8 = 4x - 8\]

\[4x(x - 3)^2 - 4x = 0\]

\[4x \big[ (x - 3)^2 - 1 \big] = 0\]

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

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Intersections: \(x = 0, \quad x = 2, \quad x = 4\)

Show the Two Areas Are Equal

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Sign check: at \(x = 1\): \(h(1) = 4 \cdot 1 \cdot (-1)(-3) = 12 > 0\) (\(f\) above \(g\))

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

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\[A_1 = \int_0^2 h(x) \, dx = \left[ x^4 - 8x^3 + 16x^2 \right]_0^2 = 16 - 64 + 64 = 16\]

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

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\[A_2 = \left| \left[ x^4 - 8x^3 + 16x^2 \right]_2^4 \right| = \left| (256 - 512 + 256) - 16 \right| = 16\]

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\[\boxed{A_1 = A_2 = 16 \quad \checkmark}\]

Walkthrough: Exponential Function - 15 Minutes

The Exponential Problem

Consider \(j(x) = (x + p) \cdot e^x - q\) with parameters \(p\), \(q\) shown below:

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

Read \(p\) and \(q\) From the Graph

Two readable points:

\(y\)-intercept: \(j(0) \approx -3\)
\(x\)-intercept: \(j(1.5) \approx 0\)

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System:

\[j(0) = p - q = -3 \quad (I)\]

\[j(1.5) = (1.5 + p) e^{1.5} - q = 0 \quad (II)\]

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Round to one decimal: \(\;\boxed{p \approx -1, \quad q \approx 2}\)

End Behavior

For \(j(x) = (x - 1) e^x - 2\):

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As \(x \to +\infty\): \((x - 1) \to \infty\) and \(e^x \to \infty\):

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

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As \(x \to -\infty\): exponential decay beats linear growth, so \((x - 1)e^x \to 0\):

\[\lim_{x \to -\infty} j(x) = 0 - 2 = -2\]

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\[\boxed{j \to +\infty \text{ at } +\infty; \quad y = -2 \text{ horizontal asymptote at } -\infty}\]

Inflection Point

For \(j(x) = (x - 1) e^x - 2\), compute:

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

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Set \(j''(x) = 0\): since \(e^x > 0\) always, \(x + 1 = 0 \Rightarrow x = -1\).

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\(y\)-coordinate:

\[j(-1) = (-1 - 1) e^{-1} - 2 = -\frac{2}{e} - 2 \approx -2.74\]

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\[\boxed{\text{Inflection point at } \left( -1, \; -\tfrac{2}{e} - 2 \right) \approx (-1, -2.74)}\]

The Scenario

A fitness studio offers a workout-tracking app to its members.

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Given:

20% of random people are members (\(M\)): \(P(M) = 0.20\)
75% of members use the app (\(A\)): \(P(A \mid M) = 0.75\)
10% of non-members still use such an app: \(P(A \mid \bar{M}) = 0.10\)

Build the Contingency Table

Use \(N = 100{,}000\) people:

\(M\): \(100{,}000 \times 0.20 = 20{,}000\); \(\bar{M}\): \(80{,}000\)
\(M \cap A\): \(20{,}000 \times 0.75 = 15{,}000\); \(M \cap \bar{A}\): \(5{,}000\)
\(\bar{M} \cap A\): \(80{,}000 \times 0.10 = 8{,}000\); \(\bar{M} \cap \bar{A}\): \(72{,}000\)

. . .

	\(M\)	\(\bar{M}\)	Sum
\(A\)	15,000	8,000	23,000
\(\bar{A}\)	5,000	72,000	77,000
Sum	20,000	80,000	100,000

Bayes: \(P(M \mid A)\)

A person uses the app. What is the probability they are a member?

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Read directly from the table:

\[P(M \mid A) = \frac{|M \cap A|}{|A|} = \frac{15{,}000}{23{,}000} \approx 0.6522\]

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\[\boxed{P(M \mid A) \approx 65.2\%}\]

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Even though only 20% of people are members, having the app raises the probability of membership to 65%, because the app is much more common among members.

Binomial: Sample of \(n = 8\)

Use \(p = P(M) = 0.20\).

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Exactly 2 are members:

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\[P(X = 2) = \binom{8}{2} (0.2)^2 (0.8)^6 = 28 \cdot 0.04 \cdot 0.2621 \approx 0.2936\]

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At most 2 are members:

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\(P(X = 0) = (0.8)^8 \approx 0.1678\)
\(P(X = 1) = 8 \cdot 0.2 \cdot (0.8)^7 \approx 0.3355\)
\(P(X = 2) \approx 0.2936\)
\(P(X \leq 2) \approx 0.7969 \approx 79.7\%\)

Binomial: Range Probability

Between 2 and 4 members (inclusive):

\(P(X = 2) \approx 0.2936\)
\(P(X = 3) = \binom{8}{3} (0.2)^3 (0.8)^5 = 56 \cdot 0.008 \cdot 0.3277 \approx 0.1468\)
\(P(X = 4) = \binom{8}{4} (0.2)^4 (0.8)^4 = 70 \cdot 0.0016 \cdot 0.4096 \approx 0.0459\)

. . .

\[\boxed{P(2 \leq X \leq 4) \approx 0.4863 \approx 48.6\%}\]

Exam Tips & Key Takeaways - 5 Minutes

What to Remember

Verify extrema in 3 steps: \(f'(a) = 0\), sign of \(f''(a)\), value \(f(a)\)
Areas between curves: find intersections; integrate the absolute difference; split at sign changes
Reading parameters: every readable point gives one equation; solve the system, then round
Contingency tables turn conditional info into counts
Binomial: identify \(n\), \(p\), and the type of event (exact, at most, at least, range)

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The tasks file for 09-06 has another exam-style problem set covering the same techniques with different numbers.