Session 09-05 - Final Recap: Calculus & Curve Sketching

Section 09: Exam Preparation

Dr. Nikolai Heinrichs & Dr. Tobias Vlćek

Concept Refresher - Part A - 25 Minutes

Domain & Range

Function type	Domain rule	Range
Polynomial \(p(x)\)	\(\mathbb{R}\)	Depends on degree & leading coeff.
Rational \(\frac{p(x)}{q(x)}\)	\(\mathbb{R} \setminus \{x : q(x) = 0\}\)	Excludes horizontal asymptote value
\(\ln(g(x))\)	\(g(x) > 0\)	\(\mathbb{R}\)
\(\sqrt{g(x)}\)	\(g(x) \geq 0\)	\([0, \infty)\)
\(e^{g(x)}\)	\(\mathbb{R}\)	\((0, \infty)\)

Key exam skill: Always check for division by zero, negative log/sqrt arguments, and context restrictions (\(t \geq 0\) for time).

Limits & Asymptotes

Horizontal asymptotes of \(\frac{a_n x^n + \ldots}{b_m x^m + \ldots}\); compare the degrees:

• \(n < m\) (numerator degree smaller): HA is \(y = 0\)
• \(n = m\) (degrees equal): HA is \(y = \frac{a_n}{b_m}\)
• \(n > m\) (numerator degree larger): no horizontal asymptote

Vertical: denominator zeros after cancelling common factors.

• Cancelled factors \(\to\) holes, not asymptotes!

Limits & Asymptotes: Example

Example: Find all asymptotes of \(f(x) = \dfrac{2x^2 - 8}{x^2 - x - 6}\).

Step 1, factor: \(f(x) = \dfrac{2(x-2)(x+2)}{(x-3)(x+2)} = \dfrac{2(x-2)}{x-3}, \quad x \neq -2\)

• Hole at \(x = -2\) (cancelled factor), with \(y = \frac{2(-4)}{-5} = \frac{8}{5}\)
• Vertical asymptote at \(x = 3\) (remaining denominator zero)
• Horizontal asymptote \(y = 2\) (degrees equal, ratio \(\frac{2}{1} = 2\))

End Behavior of Polynomials

The leading term \(a_n x^n\) determines what happens as \(x \to \pm\infty\):

• Even degree, \(a_n > 0\): \(\uparrow\) both ends \(\quad\) | \(\quad\) \(a_n < 0\): \(\downarrow\) both ends
• Odd degree, \(a_n > 0\): \(\downarrow\) left, \(\uparrow\) right \(\quad\) | \(\quad\) \(a_n < 0\): \(\uparrow\) left, \(\downarrow\) right

Example 1: \(f(x) = -3x^4 + 5x^2 - 1\)

Leading term: \(-3x^4\) \(\quad\) (even degree, \(a_n < 0\)) \(\quad \Rightarrow \quad \downarrow\) both ends

Example 2: \(f(x) = \frac{1}{8}x^3 - \frac{3}{4}x^2 + 4\)

Leading term: \(\frac{1}{8}x^3\) \(\quad\) (odd degree, \(a_n > 0\)) \(\quad \Rightarrow \quad \downarrow\) left, \(\uparrow\) right

Even degree = same direction on both sides. Odd degree = opposite directions.

Evaluating Limits

Rational functions as \(x \to \infty\): divide every term by the highest power of \(x\) in the denominator.

\[\lim_{x \to \infty} \frac{3x^2+1}{x^2-x}\]

\[\lim_{x \to \infty} \frac{3 + 1/x^2}{1 - 1/x}\]

\[\frac{3}{1} = 3\]

Derivative Rules

Product rule: \((f \cdot g)' = f' \cdot g + f \cdot g'\)

\[\text{Example: } (x^2 \ln x)' = 2x \cdot \ln x + x^2 \cdot \tfrac{1}{x} = 2x\ln x + x\]

Quotient rule: \(\left(\frac{f}{g}\right)' = \frac{f' g - f g'}{g^2}\)

\[\text{Example: } \left(\frac{x^2}{x+1}\right)' = \frac{2x(x+1) - x^2}{(x+1)^2} = \frac{x^2 + 2x}{(x+1)^2}\]

Chain rule: \((f(g(x)))' = f'(g(x)) \cdot g'(x)\)

\[\text{Example: } (e^{x^2-3x})' = e^{x^2-3x} \cdot (2x - 3)\]

Common Derivative Pitfalls

• Forgetting the chain rule: \((e^{g(x)})' = e^{g(x)} \cdot g'(x)\), not just \(e^{g(x)}\)!
• Product rule has two terms: \((f \cdot g)' \neq f' \cdot g'\)
• Sign in quotient rule: it’s \(f'g \mathbf{-} fg'\) in the numerator; order matters!
• \(\frac{d}{dx}\ln(g(x)) = \frac{g'(x)}{g(x)}\): don’t forget the \(g'(x)\) in the numerator

Trigonometric Derivatives

Property	\(\sin(x)\)	\(\cos(x)\)
Domain / Range	\(\mathbb{R}\) / \([-1, 1]\)	\(\mathbb{R}\) / \([-1, 1]\)
Period	\(2\pi\)	\(2\pi\)
Derivative	\(\cos(x)\)	\(-\sin(x)\)

With chain rule:

\((\sin(kx))' = k\cos(kx)\), \(\quad (\cos(kx))' = -k\sin(kx)\)

Angle of intersection:

A line with slope \(m\) meets the \(x\)-axis at angle \(\alpha = \arctan(m)\).

Make sure your calculator is in degree mode if the problem asks for degrees!

Tangent Lines & Curve Sketching

Tangent line at \(x = a\): \(\quad \boxed{t(x) = f'(a)(x - a) + f(a)}\)

1. Domain: where is \(f\) defined?
2. Zeros & \(y\)-intercept: \(f(x) = 0\); \(f(0)\)
3. \(f'(x) = 0\): critical points \(\to\) local max/min
4. \(f''(x) = 0\): inflection points, concavity
5. Asymptotes / end behavior
6. Sketch: combine all information

Even if only asked for the sketch, showing steps 1-5 earns partial credit!

Antiderivatives

Key rules:

• \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1)\)
• \(\int \frac{1}{x} \, dx = \ln|x| + C\)
• \(\int e^{kx} \, dx = \frac{1}{k}e^{kx} + C\)
• \(\int \sin(x) \, dx = -\cos(x) + C\), \(\quad \int \cos(x) \, dx = \sin(x) + C\)

Never forget \(+C\) on indefinite integrals!

Graphical Differentiation & Integration

Reading \(f'\) from \(f\) (graphical differentiation):

• \(f\) has local max/min \(\Rightarrow\) \(f'(x) = 0\)
• \(f\) is increasing \(\Rightarrow\) \(f' > 0\) \(\quad\) | \(\quad\) \(f\) decreasing \(\Rightarrow\) \(f' < 0\)
• If \(f\) is degree \(n\), then \(f'\) is degree \(n - 1\)

Reading \(F\) from \(f\) (graphical integration):

• \(f > 0 \Rightarrow F\) increasing \(\quad\) | \(\quad\) \(f < 0 \Rightarrow F\) decreasing
• \(f\) changes sign \(\Rightarrow\) \(F\) has a local extremum
• \(f = 0\) but no sign change (double root) \(\Rightarrow\) \(F\) has an inflection point
• If \(f\) is degree \(n\), then \(F\) is degree \(n + 1\)

Coffee Break - 15 Minutes

Exam Problem Walkthrough - 45 Minutes

The Problem

Consider the following third degree polynomial \(f(x)\):

Figure 1: Graph of the third degree polynomial f(x)

What Can We Read from This Graph?

• Zeros: \(f(-2) = 0\) and \(f(4) = 0\)
• At \(x = -2\): the graph crosses the \(x\)-axis \(\Rightarrow\) single root
• At \(x = 4\): graph touches the \(x\)-axis and turns around \(\Rightarrow\) double root
• Local maximum: at \(x = 0\) with \(f(0) = 4\)
• \(f\) is increasing for \(x < 0\), decreasing for \(0 < x < 4\), increasing for \(x > 4\)

Single vs. double root: If the graph crosses the \(x\)-axis, it’s a single root. If it touches and turns, it’s a double root (the factor appears squared).

Part a) Graph the Derivative \(f'(x)\)

Where is \(f'(x) = 0\)? At every local max/min and every double-root:

• \(f'(0) = 0\) (local maximum at \(x = 0\))
• \(f'(4) = 0\) (double root; the tangent is horizontal there)

What is the sign of \(f'\)?

• \(f\) increasing for \(x < 0\) \(\Rightarrow\) \(f' > 0\)
• \(f\) decreasing for \(0 < x < 4\) \(\Rightarrow\) \(f' < 0\)
• \(f\) increasing for \(x > 4\) \(\Rightarrow\) \(f' > 0\)
• \(f\) is cubic (degree 3) with positive leading coefficient \(\Rightarrow\) \(f'\) is a parabola opening upward

Part a) Solution

Figure 2: Graph of f(x) with its derivative f’(x)

Part b) Graph the Antiderivative \(F(x)\)

Strategy: \(F\) increases where \(f > 0\) and decreases where \(f < 0\).

• \(f < 0\) for \(x < -2\) \(\Rightarrow\) \(F\) is decreasing
• \(f > 0\) for \(-2 < x < 4\) and \(x > 4\) \(\Rightarrow\) \(F\) is increasing
• \(f(-2) = 0\) and \(f\) changes sign \(\Rightarrow\) \(F\) has a local minimum at \(x = -2\)
• \(f(4) = 0\) but \(f\) does not change sign (double root) \(\Rightarrow\) \(F\) has an inflection point at \(x = 4\)
• \(f\) is degree 3 \(\Rightarrow\) \(F\) is degree 4 (grows much faster, so we scale \(F\) by \(\frac{1}{4}\) to fit the plot)

Part b) Solution

Figure 3: Graph of f(x) with its antiderivative F(x) (scaled by 1/4 for visibility)

Part c) Strategy: Setting Up the System

We want \(f(x) = ax^3 + bx^2 + cx + d\): that’s 4 unknowns, so we need 4 equations.

What can we read from the graph?

• \(f(0) = 4\) (the \(y\)-intercept) \(\quad \Rightarrow \quad d = 4\)
• \(f'(0) = 0\) (local max; horizontal tangent) \(\quad \Rightarrow \quad c = 0\)
• \(f(-2) = 0\) (zero of the function)
• \(f(4) = 0\) (zero of the function)

Look for points where you can read exact values: zeros, \(y\)-intercept, and extrema give the cleanest equations.

Part c) Solving the System

With \(c = 0\) and \(d = 4\), the two remaining equations are:

\[f(-2) = 0: \quad -8a + 4b + 4 = 0 \quad \Rightarrow \quad 2a - b = 1 \quad (I)\]

\(f(4) = 0: \quad 64a + 16b + 4 = 0 \quad \Rightarrow \quad 64a + 16b = -4 \quad (II)\)$

\[\boxed{f(x) = \frac{1}{8}x^3 - \frac{3}{4}x^2 + 4 = \frac{1}{8}(x+2)(x-4)^2}\]

Part d) Tangent Line at \(x = 6\)

Step 1: function value at the point

• \(f(6) = \frac{216}{8} - \frac{108}{4} + 4 = 27 - 27 + 4 = 4\)

Step 2: slope (derivative at the point)

• \(f'(x) = \frac{3x^2}{8} - \frac{3x}{2} \qquad f'(6) = \frac{108}{8} - 9 = \frac{27}{2} - 9 = \frac{9}{2}\)

Step 3: plug into tangent formula \(t(x) = f'(a)(x - a) + f(a)\)

• \(t(x) = \frac{9}{2}(x - 6) + 4 = \frac{9}{2}x - 27 + 4 = \boxed{\frac{9}{2}x - 23}\)

Part e) Angle of Intersection

The tangent has slope \(m = \frac{9}{2} = 4.5\).

The angle between the tangent and the \(x\)-axis:

\[\tan(\alpha) = m = 4.5 \qquad \Rightarrow \qquad \alpha = \arctan(4.5) \approx 77.47°\]

\[\boxed{\alpha \approx 77.5°}\]

Visual

Figure 4: Graph of f(x) with tangent line at x = 6 and angle of intersection

Exam Tips & Key Takeaways - 5 Minutes

What to Remember

• Read the graph carefully before calculating; identify zeros, extrema, special points
• 4 unknowns need 4 equations: look for readable values (zeros, intercepts, extrema)
• Show your work on every step; partial credit is awarded for correct reasoning
• Verify your results: plug solutions back into the original equations
• Don’t forget the chain rule: the #1 source of derivative errors
• \(+C\) on indefinite integrals: easy points lost if forgotten
• Check your calculator mode: degrees vs. radians for \(\arctan\)

Practice with the tasks file and the mock exams, same problem structure, different numbers!