Coverage: Lessons 21-31

Exam Date and Time: Wednesday, November 18, 8:00pm - 9:00pm

Location: Announced on Brightspace

Maxima and Minima [Lesson 21]

NotesPractice

The Extreme Value Theorem

Extreme Value Theorem: If \(f\) is continuous on a closed interval \([a, b]\), then \(f\) has both an absolute maximum and an absolute minimum on \([a, b]\).

The absolute extrema (maximum and minimum) can occur at:

Critical Points

Definition: A critical point of a function \(f(x)\) is a point \(x = c\) in the domain of \(f\) where \(f'(c) = 0\), OR \(f'(c)\) does not exist (DNE).

Finding Absolute Maximum and Minimum Values

Three-Step Process:

  1. Find all critical points: compute \(f'(x)\); solve \(f'(x) = 0\); identify points where \(f'(x)\) does not exist. Only consider critical points that are within the given interval.
  2. Evaluate the function at all critical points and endpoints: calculate \(f(a)\) and \(f(b)\) for endpoints, and \(f(c)\) for each critical point \(c\).
  3. Compare all values: the largest value is the absolute maximum; the smallest value is the absolute minimum.

Common Mistakes: forgetting to check endpoints; including critical points that are outside the given interval; assuming the maximum must occur at a critical point (it could be at an endpoint!); not checking where the derivative doesn't exist; confusing the location of the extremum with its value.

Practice Problems

Rolle's Theorem & the Mean Value Theorem [Lesson 22]

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Rolle's Theorem

Rolle's Theorem: Let \(f(x)\) be a function that satisfies the following conditions:

  1. \(f(x)\) is continuous on the closed interval \([a, b]\)
  2. \(f(x)\) is differentiable on the open interval \((a, b)\) (the derivative exists)
  3. \(f(a) = f(b)\) (same endpoint values)

Then there exists at least one number \(c\) such that \(a < c < b\) and \(f'(c) = 0\) (horizontal tangent).

When Rolle's Theorem fails: Rolle's Theorem will not apply if any of the three conditions are violated. A graph with a jump discontinuity or break (not continuous), or with a sharp corner or cusp (not differentiable), may not have a point where \(f'(c) = 0\). If the endpoint values are different, this situation is handled by the Mean Value Theorem instead.

Mean Value Theorem (MVT)

Mean Value Theorem: Let \(f(x)\) be a function that satisfies:

  1. \(f(x)\) is continuous on the closed interval \([a, b]\)
  2. \(f(x)\) is differentiable on the open interval \((a, b)\)

Then there exists at least one number \(c\) such that \(a < c < b\) and \[f'(c) = \frac{f(b) - f(a)}{b - a}.\]

This means there is a point in the middle where the tangent line is parallel to the secant line connecting the endpoints.

Connection to Rolle's Theorem: Rolle's Theorem is a special case of the Mean Value Theorem where \(f(a) = f(b)\). In this case, the slope of the secant line is \(\frac{f(b) - f(a)}{b - a} = \frac{0}{b - a} = 0\), which gives us \(f'(c) = 0\).

Practice Problems

Intervals of Increasing/Decreasing & the First Derivative Test [Lesson 22]

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What the Derivative Tells Us

\(f(x)\) is increasing on an interval if \(f'(x) > 0\) on that interval; \(f(x)\) is decreasing on an interval if \(f'(x) < 0\) on that interval.

First Derivative Test

Local Maximum: A function \(f(x)\) has a relative/local maximum at \(x = c\) if \(f'(c) = 0\), \(f(x)\) is increasing on the left of \(c\) (i.e., \(f'(x) > 0\) for \(x < c\)), and \(f(x)\) is decreasing on the right of \(c\) (i.e., \(f'(x) < 0\) for \(x > c\)).

Local Minimum: A function \(f(x)\) has a relative/local minimum at \(x = c\) if \(f'(c) = 0\), \(f(x)\) is decreasing on the left of \(c\) (i.e., \(f'(x) < 0\) for \(x < c\)), and \(f(x)\) is increasing on the right of \(c\) (i.e., \(f'(x) > 0\) for \(x > c\)).

Key Takeaway: Not every critical point (where \(f'(x) = 0\)) is a local maximum or minimum. We must check the sign changes of the derivative to determine the nature of each critical point.

Practice Problems

Concavity & the Second Derivative Test [Lesson 23]

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What Does the Second Derivative Tell Us?

The second derivative \(f''(x) = \frac{d}{dx}[f'(x)]\) provides information about how the first derivative is changing. If \(f''(x) > 0\), then \(f'(x)\) is increasing; if \(f''(x) < 0\), then \(f'(x)\) is decreasing.

Concavity

Concave Up: A function is concave up on an interval where \(f''(x) > 0\). The graph curves upward, like a cup (U shape). The slope \(f'(x)\) is increasing.

Concave Down: A function is concave down on an interval where \(f''(x) < 0\). The graph curves downward, like a cap (∩ shape). The slope \(f'(x)\) is decreasing.

Second Derivative Test

For a Relative/Local Maximum at \(x = c\): \(f'(c) = 0\) and \(f''(c) < 0\) (concave down).

For a Relative/Local Minimum at \(x = c\): \(f'(c) = 0\) and \(f''(c) > 0\) (concave up).

What happens when \(f''(c) = 0\)? When the second derivative equals zero at a critical point, the Second Derivative Test is inconclusive. You must use the First Derivative Test instead. For example, \(f(x) = x^4\), \(f(x) = -x^4\), and \(f(x) = x^3\) all have \(f''(0) = 0\): the first has a local minimum at \(x = 0\), the second a local maximum, and the third neither.

Points of Inflection

Point of Inflection: A point where the concavity of the function changes. At a point of inflection, the function changes from concave up to concave down, or vice versa.

Important Note: Not every point where \(f''(x) = 0\) is necessarily a point of inflection. You must verify that the concavity actually changes on either side of the point.

Key Relationships
ConditionMeaning
\(f'(x) > 0\)Function is increasing
\(f'(x) < 0\)Function is decreasing
\(f'(x) = 0\)Horizontal tangent (potential max/min)
\(f''(x) > 0\)Concave up (slopes increasing)
\(f''(x) < 0\)Concave down (slopes decreasing)
\(f''(x) = 0\)Potential inflection point

Practice Problems

Graphing Functions [Lessons 24–25]

Notes (Lesson 24)Notes (Lesson 25)Practice

Review: What Do \(f'(x)\) and \(f''(x)\) Tell Us?

First Derivative \(f'(x)\): if \(f'(x) > 0\), then \(f(x)\) is increasing ↑; if \(f'(x) < 0\), then \(f(x)\) is decreasing ↓; if \(f'(x) = 0\) or \(f'(x)\) DNE, then we have a critical point.

Second Derivative \(f''(x)\): if \(f''(x) > 0\), then \(f(x)\) is concave up ∪; if \(f''(x) < 0\), then \(f(x)\) is concave down ∩; if \(f''(x) = 0\), then we have a potential inflection point.

Special Note on Slant Asymptotes: For a rational function \(f(x) = \frac{p(x)}{q(x)}\), if the degree of \(p(x)\) is exactly one more than the degree of \(q(x)\), then the function has a slant (oblique) asymptote. When \(x\) is large, the function approaches a linear function.

Sketching Guidelines

Complete Guidelines for Sketching \(y = f(x)\):

  1. Domain: Find all possible input values
  2. Intercepts: x-intercepts: Solve \(f(x) = 0\); y-intercept: Calculate \(y = f(0)\)
  3. Vertical Asymptotes: Find values where \(\lim_{x \to a^-} f(x) = \pm\infty\) or \(\lim_{x \to a^+} f(x) = \pm\infty\)
  4. End Behavior: Calculate \(\lim_{x \to \pm\infty} f(x)\). May reveal horizontal asymptotes, may reveal slant asymptotes, or show that \(f(x) \to \pm\infty\) (no asymptote)
  5. First Derivative \(f'(x)\): If \(f'(x) > 0\): function is increasing ↑; if \(f'(x) < 0\): function is decreasing ↓
  6. Second Derivative \(f''(x)\): If \(f''(x) > 0\): function is concave up ∪; if \(f''(x) < 0\): function is concave down ∩; if \(f''(x) = 0\): could have inflection points
  7. Symmetry: Check for even or odd functions (optional but helpful)

Symmetry

Checking for symmetry can simplify the graphing process by reducing the amount of analysis needed.

Even Function: A function \(f(x)\) is even if \(f(-x) = f(x)\) for all \(x\). Even functions have y-axis symmetry. The graph is symmetric about the y-axis, meaning if you fold it along the y-axis, both halves match perfectly. Examples: \(x^2\), \(x^4\), \(\cos(x)\), \(|x|\).

Odd Function: A function \(f(x)\) is odd if \(f(-x) = -f(x)\) for all \(x\). Odd functions have origin symmetry. The graph is symmetric about the origin, meaning if you rotate it 180° about the origin, it looks the same. Examples: \(x\), \(x^3\), \(\sin(x)\), \(\frac{1}{x}\).

When Given \(f'(x)\) Instead of \(f(x)\)

  • Use \(f'(x)\) to find where \(f(x)\) is increasing/decreasing
  • Find \(f''(x)\) by differentiating \(f'(x)\)
  • Use \(f''(x)\) to find concavity and inflection points

The sign of \(f'(x)\) tells you the direction of the graph; the sign of \(f''(x)\) tells you the curvature of the graph. Where \(f'(x) = 0\): potential extrema. Where \(f''(x) = 0\): potential inflection points.

Practice Problems

Optimization [Lessons 26–27]

Notes (Lesson 26)Notes (Lesson 27)Practice

Strategy for Optimization Problems

Four-Step Strategy:

  1. Read carefully, sketch, and label: draw a diagram if possible; label all quantities with variables; understand what's being asked.
  2. Identify the objective and constraint: Objective: What you need to maximize or minimize (look for words like "find max," "minimize," "largest," "smallest"). Constraint: Given conditions or relationships (look for phrases like "given that," "such that," "subject to").
  3. Use the constraint to write the objective as a function of one variable: express the objective function in terms of a single variable; determine the domain (valid range for the variable).
  4. Use \(f'(x)\) and \(f''(x)\) to find the maximum or minimum: find critical points: \(f'(x) = 0\); use the second derivative test or check endpoints; verify your answer makes sense in context.

Common Mistakes to Avoid: forgetting to check the domain restrictions; not checking endpoints on closed intervals; confusing the objective with the constraint; forgetting to verify that a critical point is actually a maximum or minimum; making algebra errors when simplifying the objective function; not stating the final answer in the context of the problem.

Useful Techniques

  • Similar triangles: Helpful for inscribed shapes and geometric relationships
  • Minimize \(d^2\) instead of \(d\): Simplifies distance problems
  • Second derivative test: Quickly verifies whether a critical point is a max or min
  • Check endpoints: Always consider boundary values when working on closed intervals

Note on the Distance Problem: When minimizing distance, it's often easier to minimize the square of the distance rather than the distance itself. This avoids dealing with square roots in the derivative, while still giving the same optimal point (since \(f(x)\) and \([f(x)]^2\) have their minima at the same location when \(f(x) \geq 0\)).

Common Applications: surface area minimization (packaging, containers, materials); volume maximization (storage, containers, inscribed shapes); distance problems (closest point, shortest path); cost minimization (manufacturing, construction, resources).

Practice Problems

Linear Approximation & Differentials [Lesson 28]

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Linear Approximation

Definition: Linear Approximation. The linear approximation (or tangent line approximation) of a function \(f(x)\) at a point \(x = a\) is given by: \[L(x) = f(a) + f'(a)(x - a)\] This equation represents the tangent line to \(f(x)\) at the point \((a, f(a))\), and can be used to approximate values of \(f(x)\) for \(x\) near \(a\).

Key Concept: Linear approximation works best when \(x\) is close to \(a\). The farther \(x\) is from \(a\), the less accurate the approximation becomes.

Steps for Linear Approximation

  1. Identify the function \(f(x)\) and the value you want to approximate.
  2. Choose a point \(a\) close to your target value where \(f(a)\) is easy to compute.
  3. Find the derivative \(f'(x)\) and evaluate \(f'(a)\).
  4. Write the linear approximation: \(L(x) = f(a) + f'(a)(x - a)\).
  5. Substitute your target value into \(L(x)\) to get the approximation.

Accuracy of Approximations: The accuracy of a linear approximation depends on the distance between \(x\) and \(a\) (smaller distances yield better approximations); the behavior of the function (functions with less curvature give better linear approximations); and the second derivative (if \(|f''(x)|\) is small near \(a\), the approximation is more accurate).

Error and Percentage Error

Absolute Error: \(\text{Error} = |f(x) - L(x)|\), where \(f(x)\) is the actual value and \(L(x)\) is the approximated value.

Percentage Error: \(\text{Percentage Error} = \dfrac{|f(x) - L(x)|}{|f(x)|} \times 100\%\). This expresses the error as a percentage of the actual value.

Common Mistake: Don't confuse the error in the approximation with the value \(x - a\). The error \(|f(x) - L(x)|\) measures how far the approximation is from the true value, while \(x - a\) measures the distance from the point of tangency.

Differentials

Given a function \(y = f(x)\), if \(x\) changes from \(x\) to \(x + \Delta x\), then the corresponding exact change in \(y\) is \(\Delta y = f(x + \Delta x) - f(x)\). The differential of \(x\) is defined by \(dx = \Delta x\), while the differential of \(y\) is defined by \(dy = f'(x)\,dx\). In general, \(\Delta y \neq dy\), but \(\Delta y \approx dy\).

Practice Problems

L'Hospital's Rule [Lesson 29]

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L'Hospital's Rule

L'Hospital's Rule: If \(\displaystyle \lim_{x \to a} \frac{f(x)}{g(x)}\) is of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then: \[\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}\] provided the limit on the right exists or is \(\pm\infty\).

Warning: L'Hospital's Rule can ONLY be applied when the limit is of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). Always check the form first!

Growth Rates of Functions

Hierarchy of Growth Rates: For \(b > 1\): logarithmic functions grow the slowest (\(\ln x\)); polynomial functions grow faster than logarithmic (\(x^n\) for any \(n > 0\)); exponential functions grow fastest (\(b^x\) or \(e^x\)). This gives us the following limit results as \(x \to \infty\): \[\lim_{x \to \infty} \frac{\ln x}{x^n} = 0 \quad \text{and} \quad \lim_{x \to \infty} \frac{x^n}{e^x} = 0\] for any \(n > 0\).

Other Indeterminate Forms

L'Hospital's Rule directly applies to \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\) forms. However, other indeterminate forms can be converted to these forms.

Strategy for \(0 \cdot \infty\): Convert the product to a quotient by rewriting as \(f(x) \cdot g(x) = \dfrac{f(x)}{\frac{1}{g(x)}}\) or \(f(x) \cdot g(x) = \dfrac{g(x)}{\frac{1}{f(x)}}\). This converts the form to either \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), allowing L'Hospital's Rule to be applied.

Strategy for \(\infty - \infty\): Combine terms into a single fraction to obtain \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) form.

Strategy for \(1^\infty\), \(0^0\), and \(\infty^0\): Take the natural logarithm of both sides, evaluate the limit of the logarithm (often converting to \(0 \cdot \infty\) form), then exponentiate the result.

Key Takeaway: When dealing with indeterminate forms of the type \(1^\infty\), \(0^0\), or \(\infty^0\):

  1. Set \(L\) equal to the limit
  2. Take the natural logarithm: \(\ln L = \lim[\text{logarithm of expression}]\)
  3. Simplify to get a form suitable for L'Hospital's Rule
  4. Evaluate the limit to find \(\ln L\)
  5. Exponentiate to find \(L = e^{\ln L}\)

Always verify the form is indeterminate before applying the rule; you may need to apply L'Hospital's Rule multiple times.

Practice Problems

Antiderivatives [Lesson 30]

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Antiderivative

\(F(x)\) such that \(F'(x) = f(x)\) is called antiderivative of \(f(x)\).

If \(F(x)\) & \(G(x)\) are two antiderivatives of \(f(x)\), they differ only by a constant: \(F(x) = G(x) + C\).

e.g. \(x^2,\ x^2 - 85,\ x^2 + 7,\ x^2 + C\) (any number \(C\)) are antiderivatives of \(2x\); \(-\cos x,\ -\cos x + 97,\ -\cos x + C\) are antiderivatives of \(\sin x\).

Indefinite Integral

A notation for antiderivatives: \[\int f(x)\,dx = F(x) + C\]

  • \(\int\): elongated S, means "sum" (later)
  • \(f(x)\): integrand, "the function whose antiderivative we are trying to find"
  • \(dx\): \(x\) is the variable of integration
  • \(F(x)\) is one antiderivative of \(f(x)\); \(C\) is the constant of integration

e.g. \(\int 2x\,dx = x^2 + C\), \(\quad \int \sin x\,dx = -\cos x + C\)

Antiderivatives of Some Common Functions

Idea: derivative reduces the power by 1, e.g. \(\frac{d}{dx}\left(\frac{x^4}{4}\right) = x^3\), so \(\frac{x^4}{4}\) is antiderivative of \(x^3\).

\[\begin{gathered} \int x^n\,dx = \frac{x^{n+1}}{n+1} + C,\ \ n \neq -1, \qquad \int \frac{1}{x}\,dx = \ln|x| + C\\[6pt] \int \sin x\,dx = -\cos x + C, \qquad \int \cos x\,dx = \sin x + C\\[6pt] \int \sec^2 x\,dx = \tan x + C, \qquad \int \sec x \tan x\,dx = \sec x + C, \qquad \int e^x\,dx = e^x + C\\[6pt] \int \frac{1}{1+x^2}\,dx = \tan^{-1}(x) + C, \qquad \int \frac{1}{\sqrt{1-x^2}}\,dx = \sin^{-1}(x) + C \end{gathered}\]

Derivative of right hand side is the integrand.

We know \(\left(k_1 f(x) + k_2 g(x)\right)' = k_1 f'(x) + k_2 g'(x)\), so e.g. \(\int 5x^3 + 7x^{-2}\,dx = 5\frac{x^4}{4} - 7x^{-1} + C\).

Initial Value Problems

Find \(F(x)\) such that \(F'(x) = 2x\): \(F(x) = x^2 + C\), infinitely many possible functions. If we impose an additional condition \(F(0) = 7\): \(7 = F(0) = 0 + C\), so \(C = 7\), and \(F(x) = x^2 + 7\) is the only function satisfying \(F' = 2x,\ F(0) = 7\). This is an initial value problem.

Higher order derivatives: \(\frac{d}{dx}F'(x) = F''(x)\), so \(F'(x)\) is antiderivative of \(F''(x)\). e.g. \(F''(x) = 2x\): \(F'(x) = \int 2x\,dx = x^2 + C\), \(F(x) = \int x^2 + C\,dx = \frac{x^3}{3} + Cx + D\). To find \(C, D\) we need 2 extra conditions, e.g. \(F(0) = 7,\ F'(0) = -3\) gives \(F(x) = \frac{x^3}{3} - 3x + 7\).

Practice Problems

Approximating Area: Riemann Sums [Lesson 31]

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Area Under Curves

Some areas can be computed using formulas we know. e.g. area under \(f(x) = -32x + 48\) on \([0, \frac{3}{2}]\) = area of triangle \(= \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot \frac{3}{2} \cdot 48 = 36\); area under \(y = \sqrt{1 - x^2}\) on \([-1, 1]\) = area of semicircle of radius 1 \(= \frac{1}{2}\pi(1)^2 = \frac{\pi}{2}\).

Area under the curve \(f(x) = 4x - x^2\) on \([0, 4]\) cannot be computed using formulas we know. Idea: divide into small rectangles and approximate area.

Riemann Sum

Riemann Sum approximates area under \(y = f(x)\) on \([a, b]\) using rectangles:

  1. Divide \([a, b]\) into \(n\) equal parts; each has width \(\Delta x = \dfrac{b - a}{n}\)
  2. In each small interval choose a sample point \(x_k\) & form a rectangle with height \(f(x_k)\)
  3. Approximate area by adding areas of all the rectangles
\[\text{Approx. Area using } n \text{ intervals} = f(x_1)\Delta x + f(x_2)\Delta x + \cdots + f(x_n)\Delta x\]

Notation: \(L_n\) uses left end points, \(M_n\) uses mid points, \(R_n\) uses right end points as sample points. Divide into smaller rectangles to get better approximation: for \(f(x) = 4x - x^2\) on \([0, 4]\), \(L_4 = 10\), \(M_4 = 11\), \(R_{16} = 10.625\), and the actual area is \(10.667\).

Summation Notation

\[\sum_{k=1}^{n} a_k = a_1 + a_2 + \cdots + a_n\]

Using this notation, the \(n\)th Riemann sum of \(f(x)\) on \([a, b]\), with width \(\Delta x = \frac{b - a}{n}\) and sample points \(x_1, \ldots, x_n\), is \[\text{Approx. Area} = \sum_{k=1}^{n} f(x_k)\,\Delta x\]

Some Important Summation Formulas

\[\begin{gathered} \sum_{k=1}^{n} k = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}\\[6pt] \sum_{k=1}^{n} k^2 = 1^2 + 2^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}\\[6pt] \sum_{k=1}^{n} k^3 = 1^3 + 2^3 + \cdots + n^3 = \frac{n^2(n+1)^2}{4} \end{gathered}\]

e.g. \(1 + 2 + \cdots + 10 = \frac{10(11)}{2} = 55\), \(\quad 1^2 + 2^2 + 3^2 = \frac{3(4)(7)}{6} = 14\), \(\quad 1^3 + 2^3 + 3^3 = \frac{3^2 \cdot 4^2}{4} = 36\)

Practice Problems