Session 09-06 - Final Recap: Calculus & Curve Sketching

Section 09: Exam Preparation

Dr. Nikolai Heinrichs & Dr. Tobias Vlćek

Concept Refresher

Domain & Range

Function type Domain rule Range
Polynomial \(p(x)\) \(\mathbb{R}\) By degree and sign of \(a_n\)
Rational \(\frac{p(x)}{q(x)}\) \(\mathbb{R} \setminus \{x : q(x) = 0\}\) Case by case (e.g. \(\frac{1}{x}\): \(\mathbb{R} \setminus \{0\}\))
\(\ln(g(x))\) \(g(x) > 0\) \(\mathbb{R}\) for \(\ln(x)\)
\(\sqrt{g(x)}\) \(g(x) \geq 0\) \([0; \infty)\) for \(\sqrt{x}\)
\(e^{g(x)}\) \(\mathbb{R}\) \((0; \infty)\) for \(e^x\)

Key exam skill: Check division by zero, log/sqrt arguments and context limits (\(t \geq 0\)).

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: \(f(x) = \frac{1}{12}x^3 - \frac{7}{12}x^2 + 3\)

Leading term: \(\frac{1}{12}x^3\) \(\quad\) (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

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:

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

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

Graphical Calculus

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 \(\Rightarrow\) \(f' < 0\)

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
  • \(f = 0\), no sign change (double root) \(\Rightarrow\) \(F\) has an inflection point
  • Degrees: \(f\) of degree \(n\) \(\Rightarrow\) \(f'\) has degree \(n-1\), \(F\) degree \(n+1\)

Coffee Break

Exam Problem 2

The Problem

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

Cubic curve f(x) on a unit grid: it crosses the x-axis at x = −2, 3 and 6, with a high point at (0, 3) and a low point near (4.7, −1.2).

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

What Does This Graph Tell Us?

  • Zeros: \(f(-2) = 0\), \(f(3) = 0\) and \(f(6) = 0\)
  • At all three zeros the graph crosses the \(x\)-axis \(\Rightarrow\) three single roots
  • Local maximum: at \(x = 0\) with \(f(0) = 3\)
  • Local minimum: between \(x = 4\) and \(x = 5\), not on a grid point
  • \(f\) is increasing for \(x < 0\), decreasing from \(0\) to the minimum, then increasing

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 (horizontal tangent):

  • \(f'(0) = 0\) (local maximum at \(x = 0\))
  • \(f'(x) = 0\) at the local minimum, \(x \approx 4.7\) (exactly \(\frac{14}{3}\) once we know \(f\))

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

  • Positive for \(x < 0\), negative on \((0; 4.7)\), positive for \(x > 4.7\)
  • \(f\) is cubic with positive leading coefficient \(\Rightarrow\) \(f'\) is a parabola opening upward

Part a) Solution

The cubic f(x) with its derivative, a dashed upward-opening parabola. The parabola crosses the x-axis at x = 0 and x ≈ 4.7, exactly where f has its high and low point, and is negative in between, where f falls.

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

Part b) Graph the Antiderivative \(F\)

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

  • \(f < 0\) for \(x < -2\) and on \((3; 6)\) \(\Rightarrow\) \(F\) is decreasing
  • \(f > 0\) on \((-2; 3)\) and for \(x > 6\) \(\Rightarrow\) \(F\) is increasing
  • \(f\) changes sign from \(-\) to \(+\) at \(x = -2\) and \(x = 6\) \(\Rightarrow\) \(F\) has local minima there
  • \(f\) changes sign from \(+\) to \(-\) at \(x = 3\) \(\Rightarrow\) \(F\) has a local maximum at \(x = 3\)
  • \(f\) is degree 3 \(\Rightarrow\) \(F\) is degree 4 (no double root of \(f\), so \(F\) has no saddle point)

Part b) Solution

The cubic f(x) with a dash-dotted fourth-degree curve F(x) through the origin. F has local minima at x = −2 and x = 6 and a local maximum at x = 3, about 5.4 high: exactly where f crosses the x-axis.

Figure 3: Graph of f(x) with its antiderivative F(x)

Part c) 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?

  • \(f(0) = 3\) (the \(y\)-intercept) \(\quad \Rightarrow \quad d = 3\)
  • \(f'(0) = 0\) (local max; horizontal tangent) \(\quad \Rightarrow \quad c = 0\)
  • \(f(3) = 0\) (zero of the function)
  • \(f(6) = 0\) (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 \(d = 3\), the two remaining equations are:

\[f(3) = 0: \quad 27a + 9b + 3 = 0 \quad \Rightarrow \quad 9a + 3b = -1 \quad (I)\]

\[f(6) = 0: \quad 216a + 36b + 3 = 0 \quad \Rightarrow \quad 72a + 12b = -1 \quad (II)\]

\((II) - 4 \cdot (I)\): \(\; 36a = 3 \Rightarrow a = \frac{1}{12}\). Into (I): \(\; 3b = -1 - \frac{3}{4} \Rightarrow b = -\frac{7}{12}\)

\[\boxed{f(x) = \frac{1}{12}x^3 - \frac{7}{12}x^2 + 3 = \frac{1}{12}(x+2)(x-3)(x-6)}\]

Check with the unused zero: \(f(-2) = -\frac{8}{12} - \frac{28}{12} + 3 = 0 \; \checkmark\)

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

Step 1: function value at the point

  • \(f(6) = \frac{216}{12} - \frac{7 \cdot 36}{12} + 3 = 18 - 21 + 3 = 0\) (a zero, as read from the graph)

Step 2: slope (derivative at the point)

  • \(f'(x) = \frac{x^2}{4} - \frac{7x}{6} \qquad f'(6) = \frac{36}{4} - \frac{42}{6} = 9 - 7 = 2\)

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

  • \(t(x) = 2(x - 6) + 0 = \boxed{2x - 12}\)

Part e) Angle of Intersection

The tangent has slope \(m = 2\).

The angle between the tangent and the \(x\)-axis (here also the angle at which \(f\) crosses it):

\[\tan(\alpha) = m = 2 \qquad \Rightarrow \qquad \alpha = \arctan(2) \approx 63.43°\]

\[\boxed{\alpha \approx 63.4°}\]

Visual

The cubic f(x) with its dashed tangent t(x) = 2x − 12 touching at the root x = 6. A small arc between the x-axis and the tangent marks the angle α of about 63.4°.

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

Exam Problem 1

The Problem

A producer of solar garden lamps models its costs with a third degree polynomial. Fixed costs are 448 MU. At 2 QU, total costs are 540 MU. Marginal costs at 4 QU are 36 MU. The lowest increase in costs occurs at 3 QU.

As sole supplier, the firm assumes a linear price-demand function. At 4 QU, revenue is 1040 MU and the slope of the revenue function is 240 MU.

(1 QU = 100 lamps, 1 MU = 100 Euro)

The Tasks

    1. Cost function \(C(x)\) \(\quad\) b) Revenue function \(R(x)\)
    1. Profit \(P(x)\), break-even point and profit limit
    1. Maximum profit and the price to charge
    1. Sales volume and profit at a price of 250 MU
    1. Output with lowest unit costs; long-term price floor

Part a) Translating the Conditions

\(C(x) = ax^3 + bx^2 + cx + d\), \(\quad C'(x) = 3ax^2 + 2bx + c\), \(\quad C''(x) = 6ax + 2b\)

Words in the problem Equation
Fixed costs are 448 \(C(0) = d = 448\)
At 2 QU total costs are 540 \(C(2) = 8a + 4b + 2c + 448 = 540\)
Marginal costs at 4 QU are 36 \(C'(4) = 48a + 8b + c = 36\)
Lowest increase in costs at 3 QU \(C''(3) = 18a + 2b = 0\)

“Lowest increase in costs” means the marginal cost \(C'\) is minimal, so its derivative \(C''\) is zero there.

Part a) Solving the System

From \(C''(3) = 0\): \(\quad b = -9a \quad (III)\)

Substitute into the other two (divide \(C(2)\) by 2 first):

  • \((I)\): \(\; 4a + 2b + c = 46 \; \Rightarrow \; 4a - 18a + c = 46 \; \Rightarrow \; c = 46 + 14a\)
  • \((II)\): \(\; 48a + 8b + c = 36 \; \Rightarrow \; 48a - 72a + c = 36 \; \Rightarrow \; c = 36 + 24a\)

Part a) Solving the System II

Equate both: \(\; 46 + 14a = 36 + 24a \; \Rightarrow \; a = 1, \; b = -9, \; c = 60\)

\[\boxed{C(x) = x^3 - 9x^2 + 60x + 448}\]

Check: \(C(2) = 8 - 36 + 120 + 448 = 540 \; \checkmark\), \(\quad C'(4) = 48 - 72 + 60 = 36 \; \checkmark\)

Part b) Price-Demand & Revenue

Linear price-demand \(p(x) = n - mx\) gives \(\quad R(x) = x \cdot p(x) = nx - mx^2, \quad R'(x) = n - 2mx\)

Conditions at \(x = 4\):

  • \(R(4) = 4n - 16m = 1040 \; \Rightarrow \; n - 4m = 260 \quad (I)\)
  • \(R'(4) = n - 8m = 240 \quad (II)\)

\((I) - (II)\): \(\; 4m = 20 \; \Rightarrow \; m = 5\), \(\quad n = 260 + 20 = 280\)

\[\boxed{p(x) = 280 - 5x, \qquad R(x) = 280x - 5x^2}\]

Part c) Profit Function & First Root

\[P(x) = R(x) - C(x) = 280x - 5x^2 - x^3 + 9x^2 - 60x - 448\]

\[\boxed{P(x) = -x^3 + 4x^2 + 220x - 448}\]

Break-even: \(P(x) = 0\). A cubic has no formula we can use, so test small integers:

  • \(P(1) = -1 + 4 + 220 - 448 = -225 \neq 0\)
  • \(P(2) = -8 + 16 + 440 - 448 = 0 \; \checkmark\)

So \(x = 2\) is a root and \((x - 2)\) is a factor.

Part c) Polynomial Division

\[\begin{array}{l} (-x^3 + 4x^2 + 220x - 448) : (x - 2) = -x^2 + 2x + 224 \\ \underline{-(-x^3 + 2x^2)} \\ \phantom{(-x^3}\;\; 2x^2 + 220x \\ \phantom{(-x^3}\;\; \underline{-(2x^2 - 4x)} \\ \phantom{(-x^3 + 2x^2}\;\; 224x - 448 \\ \phantom{(-x^3 + 2x^2}\;\; \underline{-(224x - 448)} \\ \phantom{(-x^3 + 2x^2 + 224x - 44}\;\; 0 \end{array}\]

Part c) The Other Break-Even Point

\[-x^2 + 2x + 224 = 0 \quad \Rightarrow \quad x^2 - 2x - 224 = 0\]

\[x = \frac{2 \pm \sqrt{4 + 896}}{2} = \frac{2 \pm 30}{2}\]

\(x = 16\) or \(x = -14\) (rejected). \(\quad \boxed{\text{Break-even: } 2 \text{ QU}, \quad \text{profit limit: } 16 \text{ QU}}\)

Part d) Maximum Profit

Necessary: \(P'(x) = -3x^2 + 8x + 220 = 0 \; \Rightarrow \; 3x^2 - 8x - 220 = 0\)

\[x = \frac{8 \pm \sqrt{64 + 2640}}{6} = \frac{8 \pm 52}{6} \; \Rightarrow \; x = 10 \; \text{ or } \; x = -\tfrac{22}{3} \; \text{(rejected)}\]

Sufficient: \(P''(x) = -6x + 8\), \(\quad P''(10) = -52 < 0 \; \checkmark\) maximum

\[P(10) = -1000 + 400 + 2200 - 448 = \boxed{1152 \text{ MU}}\]

Price to charge: \(\quad p(10) = 280 - 50 = \boxed{230 \text{ MU}}\)

Part e) Sales & Profit at Given Price

The firm asks for 250 MU. How much does it sell, and what is the profit?

Sales volume: solve \(p(x) = 250\)

\[280 - 5x = 250 \; \Rightarrow \; 5x = 30 \; \Rightarrow \; x = 6 \text{ QU} = 600 \text{ lamps}\]

Profit:

\[P(6) = -216 + 144 + 1320 - 448 = 800 \text{ MU} = \boxed{80{,}000 \text{ Euro}}\]

Only 3 points, but easy ones: solve a linear equation, plug into \(P\), convert the units.

Part f) Lowest Unit Costs

Unit costs = costs per QU:

\[\overline{C}(x) = \frac{C(x)}{x} = x^2 - 9x + 60 + \frac{448}{x}\]

Necessary: \(\overline{C}'(x) = 2x - 9 - \dfrac{448}{x^2} = 0 \quad \Big| \cdot x^2 \quad \Rightarrow \quad 2x^3 - 9x^2 - 448 = 0\)

Test \(x = 8\): \(\; 1024 - 576 - 448 = 0 \; \checkmark\)

Division: \((2x^3 - 9x^2 - 448) : (x - 8) = 2x^2 + 7x + 56\), discriminant \(49 - 448 < 0\), so no other candidates.

Part f) Long-Term Price Floor

Sufficient: \(\overline{C}''(x) = 2 + \dfrac{896}{x^3} > 0\) for all \(x > 0 \; \checkmark\) minimum

Lowest unit costs:

\[\overline{C}(8) = 64 - 72 + 60 + 56 = \boxed{108 \text{ MU per QU}}\]

Check: \(C'(8) = 192 - 144 + 60 = 108 = \overline{C}(8)\); at the minimum of the unit costs, marginal costs equal unit costs.

Long-term price floor = 108 MU. Below this price the firm cannot cover its total costs in the long run, even at its most efficient output of 8 QU.

Visual

Cost C(x), revenue R(x) and dashed profit P(x) for 0 to 20 QU. Revenue lies above cost between x = 2 and x = 16, the break-even points where the profit curve crosses zero; profit peaks at 1152 MU at x = 10.

Figure 5: Cost, revenue and profit with break-even points and maximum profit

Exam Tip: Where Points Are (a-c)

Part Pts What earns method points
a) Cost function 11 The four equations written out, \(C''(3) = 0\) named, \(a, b, c, d\)
b) Revenue 4 \(R = x \cdot p(x)\), both conditions, \(m\) and \(n\)
c) Break-even 6 \(P = R - C\), tested root, division shown, quadratic solved

Exam Tip: Where Points Are (d-f)

Part Pts What earns method points
d) Max profit 8 \(P' = 0\), \(P'' < 0\) checked, \(P_{\max}\), price \(p(x^*)\)
e) Given price 3 \(p(x) = p_0\) solved, profit with units
f) Unit costs 5 \(\overline{C} = C/x\), \(\overline{C}' = 0\), minimum checked, floor interpreted

Exam Tip: Start With the Table

Part a) is worth almost a third of the problem: write the translation table first. A wrong \(C(x)\) still earns method points in c) to f) if you carry it through correctly.

Tips & Takeaways

What to Remember

  • Read the graph first: zeros, extrema, special points
  • 4 unknowns need 4 equations - use readable values
  • Show your work - partial credit for correct reasoning
  • Verify: plug solutions back in
  • Chain rule - the #1 source of derivative errors
  • \(+C\) on indefinite integrals - easy points lost
  • Calculator mode: degrees vs. radians for \(\arctan\)

What to Remember: Problem 1

  • Translate words: fixed costs \(\to C(0)\), marginal \(\to C'\), lowest increase \(\to C'' = 0\)
  • Break-even: test small integers, divide by \((x - r)\), solve the quadratic
  • Price floor: minimum of \(\overline{C} = C/x\), where \(\overline{C} = C'\)

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