5.2

Increasing & Decreasing Functions (Monotonicity)

  • Strictly Increasing Function: A function $f(x)$ is increasing on an interval if for all $x_1 < x_2$, $f(x_1) ≤ f(x_2)$. In calculus terms: $$f'(x) > 0 \implies f(x) \text{ is strictly increasing}$$
  • Strictly Decreasing Function: A function $f(x)$ is decreasing on an interval if: $$f'(x) < 0 \implies f(x) \text{ is strictly decreasing}$$
  • Determining Monotonic Intervals:
    1. Calculate the derivative function $f'(x)$.
    2. Solve $f'(x) = 0$ and identify any points where $f'(x)$ is undefined to find critical values.
    3. Place critical values on a number line (sign diagram) to partition the domain into test intervals.
    4. Evaluate the algebraic sign of $f'(x)$ within each interval.

Example 1

Finding Intervals of Monotonicity

Find the intervals on which the function $f(x) = 2x^3 - 9x^2 + 12x - 5$ is strictly increasing and strictly decreasing.

  1. Differentiate $f(x)$: $$f'(x) = 6x^2 - 18x + 12 = 6(x^2 - 3x + 2) = 6(x - 1)(x - 2)$$
  2. Find critical values ($f'(x) = 0$): $$6(x - 1)(x - 2) = 0 \implies x = 1 \quad \text{or} \quad x = 2$$
  3. Sign test on intervals:
    • For $x < 1$ (e.g. $x = 0$): $f'(0) = 6(-1)(-2) = 12 > 0 \implies \text{Increasing}$
    • For $1 < x < 2$ (e.g. $x = 1.5$): $f'(1.5) = 6(0.5)(-0.5) = -1.5 < 0 \implies \text{Decreasing}$
    • For $x > 2$ (e.g. $x = 3$): $f'(3) = 6(2)(1) = 12 > 0 \implies \text{Increasing}$

Answer: Increasing on $x ≤ 1$ and $x ≥ 2$ (or $(-\infty, 1] \cup [2, \infty)$); Decreasing on $1 ≤ x ≤ 2$ (or $[1, 2]$).

5.2, 5.7

Stationary Points & The First Derivative Test

  • Stationary Point Definition: A point $(c, f(c))$ on the curve is stationary if the tangent line is horizontal: $$f'(c) = 0$$
  • Classification via Sign Change of $f'(x)$ (First Derivative Test):
    • Local Maximum: $f'(x)$ changes sign from positive ($+$) to negative ($-$) through $x = c$. The curve rises, levels off, then falls.
    • Local Minimum: $f'(x)$ changes sign from negative ($-$) to positive ($+$) through $x = c$. The curve falls, levels off, then rises.
    • Stationary Point of Inflection: $f'(x)$ has the same sign on both sides of $x = c$ (either $+ \to 0 \to +$ or $- \to 0 \to -$).
Classification of Stationary Points via First Derivative Illustrates three stationary behaviors: Local Maximum with plus to minus gradient change, Local Minimum with minus to plus change, and Stationary Inflection with positive gradient on both sides. Local Maximum f' > 0 f'(c) = 0 f' < 0 + 0 − (Crest) Local Minimum f' < 0 f'(c) = 0 f' > 0 − 0 + (Trough) Stationary Inflection f' > 0 f'(c) = 0 f' > 0 + 0 + (Chair)
Figure 5.3: Classification of stationary points by tracking the sign transition of the first derivative $f'(x)$.

Example 2

Classifying Stationary Points with Sign Diagrams

Find the coordinates of all stationary points on the curve $y = x^4 - 4x^3$ and classify each using the first derivative test.

  1. Differentiate $y$: $$\frac{dy}{dx} = 4x^3 - 12x^2 = 4x^2(x - 3)$$
  2. Set $\frac{dy}{dx} = 0$: $$4x^2(x - 3) = 0 \implies x = 0 \quad \text{or} \quad x = 3$$
  3. Evaluate coordinates: $$\text{At } x = 0: \quad y = (0)^4 - 4(0)^3 = 0 \implies (0, 0)$$ $$\text{At } x = 3: \quad y = (3)^4 - 4(3)^3 = 81 - 108 = -27 \implies (3, -27)$$
  4. Test sign of $f'(x)$:
    • For $x < 0$ (e.g. $-1$): $4(-1)^2(-1 - 3) = 4(-4) = -16 < 0$
    • For $0 < x < 3$ (e.g. $1$): $4(1)^2(1 - 3) = 4(-2) = -8 < 0$
    • For $x > 3$ (e.g. $4$): $4(16)(1) = 64 > 0$
    Through $x = 0$, $f'(x)$ goes $- \to 0 \to - \implies (0, 0)$ is a stationary point of inflection.
    Through $x = 3$, $f'(x)$ goes $- \to 0 \to + \implies (3, -27)$ is a local minimum.

Answers: Stationary inflection at $(0, 0)$; Local minimum at $(3, -27)$.

5.7

The Second Derivative & Concavity Test

  • Second Derivative Notation: $f''(x) = \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)$ measures the rate of change of the gradient.
  • Concavity Criteria: $$f''(x) > 0 \iff \text{Concave Upward } (\cup \text{, tangent lies below the curve})$$ $$f''(x) < 0 \iff \text{Concave Downward } (\cap \text{, tangent lies above the curve})$$
  • Second Derivative Test for Stationary Points (where $f'(c) = 0$):
    • If $f''(c) < 0$, the curve is concave down at the stationary point $\implies \mathbf{Local\ Maximum}$.
    • If $f''(c) > 0$, the curve is concave up at the stationary point $\implies \mathbf{Local\ Minimum}$.
    • If $f''(c) = 0$, the test is inconclusive; you must use the First Derivative Test.

Example 3

Applying the Second Derivative Test

For the curve $f(x) = x^3 - 6x^2 + 9x + 2$:
(a) Find $f'(x)$ and $f''(x)$.
(b) Find the stationary points and classify each using $f''(x)$.

  1. (a) Derivatives: $$f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x - 1)(x - 3)$$ $$f''(x) = 6x - 12$$
  2. (b) Stationary points ($f'(x) = 0$): $$x = 1 \quad \text{and} \quad x = 3$$ $$f(1) = 1 - 6 + 9 + 2 = 6 \implies (1, 6)$$ $$f(3) = 27 - 54 + 27 + 2 = 2 \implies (3, 2)$$
  3. Classify with $f''(x)$: $$\text{At } x = 1: \quad f''(1) = 6(1) - 12 = -6 < 0 \implies \mathbf{Local\ Maximum\ at\ (1, 6)}$$ $$\text{At } x = 3: \quad f''(3) = 6(3) - 12 = +6 > 0 \implies \mathbf{Local\ Minimum\ at\ (3, 2)}$$

Answers: Local maximum at $(1, 6)$; Local minimum at $(3, 2)$.

5.7

Points of Inflection

  • Definition of Inflection Point: A point on the curve where the concavity changes (from concave up to concave down, or vice versa). At an inflection point, the tangent line crosses through the curve.
  • Necessary Condition vs Sufficiency: $$f''(c) = 0 \quad (\text{or } f''(c) \text{ undefined}) \quad \mathbf{AND} \quad f''(x) \text{ changes sign across } x = c$$ Common mistake: $f''(c) = 0$ alone does NOT guarantee an inflection point (e.g. $f(x) = x^4$ at $x = 0$ has $f''(0) = 0$, but $f'' \ge 0$ everywhere, so concavity never changes).
  • Stationary vs Non-Stationary Inflection: $$\text{Stationary Inflection: } f'(c) = 0 \text{ and } f''(c) = 0 \text{ (with concavity sign change)}$$ $$\text{Non-Stationary Inflection: } f'(c) \ne 0 \text{ and } f''(c) = 0 \text{ (with concavity sign change)}$$
Point of Inflection and Tangent Line Crossing Illustrates a curve transitioning from concave downward f''(x) < 0 to concave upward f''(x) > 0 at a point of inflection where the tangent line crosses through the curve. x y Point of Inflection (c, f(c)) f''(c) = 0 (Concavity changes) Concave Down (∩) f''(x) < 0 Concave Up (∪) f''(x) > 0
Figure 5.4: At an inflection point, concavity switches sign and the tangent line cuts across the curve.

Example 4

Finding and Verifying a Point of Inflection

Find the coordinates of the point of inflection for $f(x) = x^3 - 3x^2 - 9x + 4$, and determine whether the tangent at that point is stationary or non-stationary.

  1. Find derivatives: $$f'(x) = 3x^2 - 6x - 9$$ $$f''(x) = 6x - 6$$
  2. Set $f''(x) = 0$: $$6x - 6 = 0 \implies x = 1$$
  3. Check sign change of $f''(x)$ across $x = 1$:
    • For $x < 1$ (e.g. $0$): $f''(0) = -6 < 0 \implies \text{Concave down}$
    • For $x > 1$ (e.g. $2$): $f''(2) = +6 > 0 \implies \text{Concave up}$
    Because $f''(x)$ changes sign from negative to positive, an inflection point exists at $x = 1$.
  4. Find point coordinates & tangent gradient: $$y = f(1) = (1)^3 - 3(1)^2 - 9(1) + 4 = 1 - 3 - 9 + 4 = -7 \implies \mathbf{(1, -7)}$$ $$f'(1) = 3(1)^2 - 6(1) - 9 = 3 - 6 - 9 = -12 \ne 0$$ Since $f'(1) \ne 0$, the tangent has gradient $m = -12$ (non-stationary).

Answers: Point of inflection is $(1, -7)$; it is a non-stationary point of inflection with tangent gradient $-12$.

5.8

Applied Optimization Modeling

  • 5-Step IB Optimization Protocol:
    1. Define Variables: Label dimensions clearly (e.g. $x$ width, $y$ length, $V$ volume).
    2. Write Equations: Formulate the objective function (quantity to maximize/minimize) and the constraint equation.
    3. Single-Variable Reduction: Rearrange the constraint to express one variable in terms of the other and substitute into the objective function.
    4. Differentiate & Solve: Set $\frac{d(\text{Objective})}{dx} = 0$ to find critical values within the physical domain.
    5. Verify Extrema: Use the Second Derivative Test ($\frac{d^2(\text{Obj})}{dx^2} < 0$ for max, $> 0$ for min) to mathematically justify the optimum.

Example 5

Classic Optimization: Maximum Rectangular Enclosure with Fixed Perimeter

A farmer has $120\text{ metres}$ of fencing to enclose a rectangular pen partitioned down the middle with an internal dividing fence parallel to one of the sides.
(a) Express the total area $A$ in terms of $x$ (the length of the dividing fence).
(b) Find the dimensions that maximize the enclosed area.
(c) Justify that this area is a maximum.

  1. (a) Constraint and Area formula: Let the dividing fence and the two parallel side fences each have length $x$ (3 pieces of length $x$).
    Let the perpendicular sides have length $y$ (2 pieces of length $y$).
    Total perimeter fencing: $$3x + 2y = 120 \implies 2y = 120 - 3x \implies y = 60 - 1.5x$$ Total area: $$A = x \cdot y = x(60 - 1.5x) = \mathbf{60x - 1.5x^2}$$
  2. (b) Maximize $A(x)$: $$\frac{dA}{dx} = 60 - 3x$$ Set $\frac{dA}{dx} = 0 \implies 60 - 3x = 0 \implies x = 20\text{ m}$ Find $y$: $$y = 60 - 1.5(20) = 60 - 30 = 30\text{ m}$$ Maximum Area: $$A_{\max} = 20 \times 30 = 600\text{ m}^2$$
  3. (c) Justification via Second Derivative: $$\frac{d^2A}{dx^2} = \frac{d}{dx}(60 - 3x) = -3$$ Since $\frac{d^2A}{dx^2} = -3 < 0$ for all $x$, the critical value $x = 20$ yields an absolute maximum area.

Answers: (a) $A(x) = 60x - 1.5x^2$; (b) Dimensions: $20\text{ m} \times 30\text{ m}$ (Area $= 600\text{ m}^2$); (c) $\frac{d^2A}{dx^2} = -3 < 0$ confirms a maximum.

Tier 1: Core Monotonicity & Extrema Fluency (Questions 1–4)

  1. 1. Intervals of Increase: For the function $f(x) = x^3 - 12x + 7$, find the set of values of $x$ for which $f(x)$ is strictly decreasing.

    Show solution

    Step 1: Find the first derivative: $$f'(x) = 3x^2 - 12 = 3(x^2 - 4) = 3(x - 2)(x + 2)$$ Step 2: Set $f'(x) < 0$ for decreasing behavior: $$3(x - 2)(x + 2) < 0 \implies -2 < x < 2$$ Conclusion: The function is strictly decreasing on the interval $\mathbf{-2 ≤ x ≤ 2}$ (or $(-2, 2)$).

  2. 2. Stationary Point Classification: Find and classify all stationary points on the curve $y = 3x^4 - 4x^3 + 1$.

    Show solution

    Step 1: Differentiate: $$\frac{dy}{dx} = 12x^3 - 12x^2 = 12x^2(x - 1)$$ Step 2: Set $\frac{dy}{dx} = 0 \implies x = 0 \text{ or } x = 1$.
    Find coordinates: $$\text{At } x = 0: \quad y = 1 \implies (0, 1)$$ $$\text{At } x = 1: \quad y = 3(1) - 4(1) + 1 = 0 \implies (1, 0)$$ Step 3: Classify with first derivative sign testing: $$\text{For } x < 0: \quad 12(+)(-) < 0$$ $$\text{For } 0 < x < 1: \quad 12(+)(-) < 0$$ $$\text{For } x > 1: \quad 12(+)(+) > 0$$ Through $x = 0$, sign is $- \to 0 \to - \implies \mathbf{(0, 1)\text{ is a stationary point of inflection}}$.
    Through $x = 1$, sign is $- \to 0 \to + \implies \mathbf{(1, 0)\text{ is a local minimum}}$.

  3. 3. Second Derivative Concavity: Determine the interval on which the curve $y = x^4 - 6x^2 + 8x - 1$ is concave down.

    Show solution

    Step 1: Compute first and second derivatives: $$\frac{dy}{dx} = 4x^3 - 12x + 8$$ $$\frac{d^2y}{dx^2} = 12x^2 - 12 = 12(x^2 - 1) = 12(x - 1)(x + 1)$$ Step 2: Concave down requires $\frac{d^2y}{dx^2} < 0$: $$12(x - 1)(x + 1) < 0 \implies \mathbf{-1 < x < 1}$$

  4. 4. Points of Inflection: Find the point of inflection of $f(x) = 2x^3 - 9x^2 + 12x - 3$ and find the gradient of the tangent at this point.

    Show solution

    Step 1: Find derivatives: $$f'(x) = 6x^2 - 18x + 12$$ $$f''(x) = 12x - 18$$ Step 2: Solve $f''(x) = 0$: $$12x - 18 = 0 \implies x = 1.5$$ $f''(x)$ changes from negative ($x < 1.5$) to positive ($x > 1.5$), so an inflection point exists.
    Step 3: Coordinates and tangent gradient: $$y = f(1.5) = 2(3.375) - 9(2.25) + 12(1.5) - 3 = 6.75 - 20.25 + 18 - 3 = 1.5 \implies \mathbf{(1.5, 1.5)}$$ $$\text{Tangent gradient } m = f'(1.5) = 6(1.5)^2 - 18(1.5) + 12 = 13.5 - 27 + 12 = \mathbf{-1.5}$$

Tier 2: IB Examination Modeling & Unknown Parameters (Questions 5–7)

  1. 5. Determining Unknown Coefficients from Extrema: The curve $y = ax^3 + bx^2 - 12x + 7$ has a stationary point at $(2, -9)$. Find the values of constants $a$ and $b$, and classify the stationary point at $x = 2$.

    Show solution

    Condition 1 (Curve passes through $(2, -9)$): $$-9 = a(2)^3 + b(2)^2 - 12(2) + 7 \implies 8a + 4b - 24 + 7 = -9 \implies 8a + 4b = 8 \implies 2a + b = 2 \quad \text{--- (1)}$$ Condition 2 (Stationary at $x = 2 \implies \frac{dy}{dx} = 0$): $$\frac{dy}{dx} = 3ax^2 + 2bx - 12$$ $$\text{At } x = 2: \quad 3a(4) + 2b(2) - 12 = 0 \implies 12a + 4b = 12 \implies 3a + b = 3 \quad \text{--- (2)}$$ Subtract (1) from (2): $$(3a + b) - (2a + b) = 3 - 2 \implies \mathbf{a = 1}$$ Substitute $a = 1$ into (1): $$2(1) + b = 2 \implies \mathbf{b = 0}$$ The equation is $y = x^3 - 12x + 7$.
    Classification: $$\frac{d^2y}{dx^2} = 6x \implies \text{At } x = 2: \quad \frac{d^2y}{dx^2} = 6(2) = 12 > 0 \implies \mathbf{Local\ Minimum}$$

  2. 6. Inflection Point on a Trigonometric Curve: Consider the function $f(x) = \sin(2x) + x$ for $0 \le x \le \pi$. Find the coordinates of the points of inflection on this interval.

    Show solution

    Step 1: Find derivatives: $$f'(x) = 2\cos(2x) + 1$$ $$f''(x) = -4\sin(2x)$$ Step 2: Solve $f''(x) = 0$ on $0 \le x \le \pi$: $$-4\sin(2x) = 0 \implies \sin(2x) = 0 \implies 2x = 0, \pi, 2\pi \implies x = 0, \frac{\pi}{2}, \pi$$ Interior candidate is $x = \frac{\pi}{2}$.
    Step 3: Test concavity change: For $x < \frac{\pi}{2}$ (e.g. $\frac{\pi}{4}$), $f''(\frac{\pi}{4}) = -4(1) = -4 < 0$.
    For $x > \frac{\pi}{2}$ (e.g. $\frac{3\pi}{4}$), $f''(\frac{3\pi}{4}) = -4(-1) = +4 > 0$.
    Concavity changes from down to up $\implies$ Inflection point at $x = \frac{\pi}{2}$.
    $$y = f\left(\frac{\pi}{2}\right) = \sin(\pi) + \frac{\pi}{2} = 0 + \frac{\pi}{2} = \frac{\pi}{2} \implies \mathbf{\left(\frac{\pi}{2}, \frac{\pi}{2}\right)}$$

  3. 7. Horizontal Tangent to a Rational Curve: For $f(x) = \frac{x^2 + 9}{x}$ ($x \ne 0$), find the coordinates of the stationary points and use the second derivative test to determine their nature.

    Show solution

    Step 1: Rewrite before differentiating: $$f(x) = x + 9x^{-1}$$ $$f'(x) = 1 - 9x^{-2} = 1 - \frac{9}{x^2}$$ $$f''(x) = \frac{d}{dx}(-9x^{-2}) = 18x^{-3} = \frac{18}{x^3}$$ Step 2: Set $f'(x) = 0$: $$1 - \frac{9}{x^2} = 0 \implies x^2 = 9 \implies x = 3 \quad \text{or} \quad x = -3$$ Coordinates: $$\text{At } x = 3: \quad y = \frac{3^2 + 9}{3} = \frac{18}{3} = 6 \implies (3, 6)$$ $$\text{At } x = -3: \quad y = \frac{(-3)^2 + 9}{-3} = \frac{18}{-3} = -6 \implies (-3, -6)$$ Step 3: Second derivative test: $$f''(3) = \frac{18}{27} > 0 \implies \mathbf{Local\ Minimum\ at\ (3, 6)}$$ $$f''(-3) = \frac{18}{-27} < 0 \implies \mathbf{Local\ Maximum\ at\ (-3, -6)}$$

Tier 3: Applied Optimization & Examination Contexts (Questions 8–10)

  1. 8. Open Box Volume Optimization: A rectangular sheet of metal measures $30\text{ cm}$ by $16\text{ cm}$. Equal squares of side length $x\text{ cm}$ are cut from each of the four corners, and the flaps are folded up to form an open rectangular box.
    (a) Show that the volume $V(x)$ is given by $V(x) = 4x^3 - 92x^2 + 480x$.
    (b) State the practical domain of $x$.
    (c) Find the value of $x$ that maximizes the volume, and calculate this maximum volume.

    Show solution

    (a) Dimensions of folded box: $$\text{Height} = x, \quad \text{Length} = 30 - 2x, \quad \text{Width} = 16 - 2x$$ $$V(x) = x(30 - 2x)(16 - 2x) = x(480 - 60x - 32x + 4x^2) = x(4x^2 - 92x + 480) = \mathbf{4x^3 - 92x^2 + 480x}$$ (b) Dimensions must be positive: $x > 0$, $30 - 2x > 0 \implies x < 15$, $16 - 2x > 0 \implies x < 8$.
    Practical domain: $\mathbf{0 < x < 8}$.

    (c) Maximize: $$\frac{dV}{dx} = 12x^2 - 184x + 480 = 0 \implies 4(3x^2 - 46x + 120) = 0$$ $$(3x - 10)(x - 12) = 0 \implies x = \frac{10}{3}\text{ cm} \quad (\approx 3.33) \quad \text{or} \quad x = 12\text{ cm}$$ Since $x = 12$ is outside the practical domain, $x = \frac{10}{3}\text{ cm}$.
    Verify with second derivative: $$\frac{d^2V}{dx^2} = 24x - 184 \implies \text{At } x = \frac{10}{3}: \quad 24\left(\frac{10}{3}\right) - 184 = 80 - 184 = -104 < 0 \implies \mathbf{Maximum}$$ Maximum volume: $$V\left(\frac{10}{3}\right) = \frac{10}{3}\left(30 - \frac{20}{3}\right)\left(16 - \frac{20}{3}\right) = \frac{10}{3}\left(\frac{70}{3}\right)\left(\frac{28}{3}\right) = \frac{19600}{27} \approx \mathbf{725.9\text{ cm}^3}$$

  2. 9. Enclosed Paddock Against a River: A farmer wishes to build a rectangular field using a straight riverbank as one of its natural boundaries (no fence needed along the river). The remaining three sides require $240\text{ m}$ of fencing.
    (a) Find the dimensions that maximize the area enclosed.
    (b) Calculate the maximum area.

    Show solution

    (a) Let $x$ be the length of the two sides perpendicular to the riverbank, and $y$ be the side parallel to the riverbank.
    Perimeter constraint: $$2x + y = 240 \implies y = 240 - 2x$$ Area function: $$A(x) = x \cdot y = x(240 - 2x) = 240x - 2x^2$$ Differentiate: $$\frac{dA}{dx} = 240 - 4x = 0 \implies x = 60\text{ m}$$ $$y = 240 - 2(60) = 120\text{ m}$$ Verify: $\frac{d^2A}{dx^2} = -4 < 0 \implies \text{Maximum}$.
    Dimensions: $\mathbf{60\text{ m} \times 120\text{ m}}$.

    (b) Maximum area: $$A_{\max} = 60 \times 120 = \mathbf{7200\text{ m}^2}$$

  3. 10. Cylindrical Can Material Minimization: A closed cylindrical drink can is to hold a volume of $500\text{ cm}^3$.
    (a) Show that the total surface area $A$ in terms of radius $r$ is $A(r) = 2\pi r^2 + \frac{1000}{r}$.
    (b) Find the exact value of the radius $r$ that minimizes the surface area.
    (c) Show that for this minimum can, the height is equal to the diameter.

    Show solution

    (a) Volume constraint: $$V = \pi r^2 h = 500 \implies h = \frac{500}{\pi r^2}$$ Total surface area (two circular ends + curved wall): $$A = 2\pi r^2 + 2\pi r h = 2\pi r^2 + 2\pi r \left(\frac{500}{\pi r^2}\right) = \mathbf{2\pi r^2 + \frac{1000}{r}}$$ (b) Minimize $A(r) = 2\pi r^2 + 1000r^{-1}$: $$\frac{dA}{dr} = 4\pi r - \frac{1000}{r^2} = 0 \implies 4\pi r = \frac{1000}{r^2} \implies r^3 = \frac{1000}{4\pi} = \frac{250}{\pi}$$ $$\mathbf{r = \sqrt[3]{\frac{250}{\pi}}\text{ cm}} \approx 4.30\text{ cm}$$ Verify: $\frac{d^2A}{dr^2} = 4\pi + \frac{2000}{r^3} = 4\pi + \frac{2000}{250/\pi} = 4\pi + 8\pi = 12\pi > 0 \implies \text{Minimum}$.

    (c) Calculate height $h$: $$h = \frac{500}{\pi r^2} = \frac{2(250)}{\pi r^2} = \frac{2(\pi r^3)}{\pi r^2} = 2r$$ Since $h = 2r = \text{diameter}$, the height equals the diameter.

Increasing / Decreasing Condition

Memorize
$$f'(x) > 0 \implies \text{Inc}, \quad f'(x) < 0 \implies \text{Dec}$$

Sign of first derivative determines whether graph is rising or falling.

Stationary Point Condition

Memorize
$$f'(x) = 0$$

Horizontal tangent gradient at critical turning values.

Second Derivative Extrema Test

Memorize
$$f''(c) < 0 \implies \text{Local Max}, \quad f''(c) > 0 \implies \text{Local Min}$$

Evaluated at stationary points where $f'(c) = 0$. Inconclusive if $f''(c) = 0$.

Point of Inflection Condition

Memorize
$$f''(c) = 0 \quad \text{AND} \quad f''(x) \text{ changes sign}$$

Concavity changes sign across $x = c$. Tangent line passes through curve.

Concavity Tests

Memorize
$$f''(x) > 0 \implies \text{Concave Up } (\cup), \quad f''(x) < 0 \implies \text{Concave Down } (\cap)$$

Determines whether tangents lie below or above the function graph.

Optimization Extrema Protocol

Memorize
$$\frac{d(\text{Obj})}{dx} = 0 \implies \text{Critical}, \quad \text{Check } \frac{d^2(\text{Obj})}{dx^2}$$

Always state physical domain and mathematically justify max/min nature.

Critical IB Examination Pitfalls

  • Assuming $f''(c) = 0$ guarantees an inflection point: You must demonstrate an actual change of sign in $f''(x)$ (e.g. $y = x^4$ has $f''(0) = 0$ but is NOT an inflection point).
  • Confusing the function value with the derivative: Coordinates of maximum/minimum points require substituting the solved $x$-value into $y = f(x)$, NOT into $f'(x)$ or $f''(x)$.
  • Forgetting to justify in Optimization: IB markschemes award an explicit mark ($M1$ or $R1$) for verifying that the critical value actually yields a maximum or minimum (via the second derivative test or sign test).
  • Omitting practical domain constraints: In physical geometry problems, lengths and radii must satisfy $x > 0$ and any container boundary cut-offs.

Section A: Diagnostic Concept Check

Instant feedback questions to test monotonicity, concavity, and stationary classifications.

Score: 0 / 4

1. If $f'(x) = 3(x - 2)(x + 4)$, on which of the following intervals is $f(x)$ strictly decreasing?

Correct! $f'(x) < 0$ when the quadratic factor $(x - 2)(x + 4) < 0$, which occurs between the roots: $-4 < x < 2$.

2. At a stationary point $x = 3$, $f'(3) = 0$ and $f''(3) = -8$. What is the nature of the stationary point?

Correct! By the second derivative test, $f''(3) < 0$ indicates the curve is concave down ($\cap$), which corresponds to a local maximum.

3. Which statement is both necessary AND sufficient for $(c, f(c))$ to be a point of inflection?

Correct! An inflection point is defined by a change in concavity, which requires $f''(x)$ to change sign across $x = c$. $f''(c) = 0$ alone is not sufficient.

4. The total cost $C(x) = 2x^2 + \frac{5000}{x}$ for $x > 0$. At what value of $x$ is the cost minimized?

Correct! $C'(x) = 4x - \frac{5000}{x^2} = 0 \implies 4x^3 = 5000 \implies x^3 = 1250 \implies x = \sqrt[3]{1250}$.

Section B: Examination Practice

Multi-step questions with markscheme breakdowns.

Paper 1 (Non-Calculator) [7 marks]

Consider the function $f(x) = x^3 - 6x^2 + 9x - 1$.

(a) Find the coordinates of the two stationary points on the graph of $f$. [3 marks]
(b) Using the second derivative, determine the nature of each stationary point. [2 marks]
(c) Find the coordinates of the point of inflection. [2 marks]
View Markscheme
(a)
M1 for setting $f'(x) = 0$: $3x^2 - 12x + 9 = 0 \implies 3(x - 1)(x - 3) = 0$
$x = 1 \implies f(1) = 1 - 6 + 9 - 1 = 3 \implies (1, 3)$ A1
$x = 3 \implies f(3) = 27 - 54 + 27 - 1 = -1 \implies (3, -1)$ A1

(b)
M1 for finding $f''(x) = 6x - 12$ and substituting values:
$f''(1) = 6(1) - 12 = -6 < 0 \implies (1, 3)\text{ is a local maximum}$ A1
$f''(3) = 6(3) - 12 = +6 > 0 \implies (3, -1)\text{ is a local minimum}$

(c)
M1 for setting $f''(x) = 0$: $6x - 12 = 0 \implies x = 2$
$f(2) = 8 - 24 + 18 - 1 = 1 \implies \mathbf{(2, 1)}$ A1
Paper 2 (Calculator Allowed) [7 marks]

A company produces cylindrical water storage tanks with closed tops. Each tank has radius $r$ metres and height $h$ metres, and must have a capacity of $100\text{ m}^3$.

(a) Show that the total surface area $A$ of the tank is given by $A = 2\pi r^2 + \frac{200}{r}$. [2 marks]
(b) Find the value of $r$ that minimizes the surface area. [3 marks]
(c) Justify that your value of $r$ gives a minimum area, and calculate the minimum area to 3 significant figures. [2 marks]
View Markscheme
(a)
$V = \pi r^2 h = 100 \implies h = \frac{100}{\pi r^2}$ M1
$A = 2\pi r^2 + 2\pi rh = 2\pi r^2 + 2\pi r\left(\frac{100}{\pi r^2}\right) = 2\pi r^2 + \frac{200}{r}$ A1

(b)
M1 for differentiating: $\frac{dA}{dr} = 4\pi r - \frac{200}{r^2}$
M1 for equating to zero: $4\pi r^3 = 200 \implies r^3 = \frac{50}{\pi}$
$r = \sqrt[3]{\frac{50}{\pi}} \approx 2.52\text{ m}$ A1

(c)
R1 for checking second derivative: $\frac{d^2A}{dr^2} = 4\pi + \frac{400}{r^3} = 4\pi + \frac{400}{50/\pi} = 12\pi > 0 \implies \text{Minimum}$
$A_{\min} = 2\pi(2.515)^2 + \frac{200}{2.515} \approx \mathbf{119\text{ m}^2}$ A1

Calculus Unit

Next Lessons in Calculus

5.1, 5.3 Limits and First Derivatives Rates of change, first principles limits, and power rule for polynomials. 5.2, 5.7, 5.8 Increasing, Decreasing & Stationary Points Monotonicity tests, second derivatives, local extrema, and inflection points. 5.4, 5.6 Tangents, Normals & Rules Product, quotient, and chain rules with tangent and normal line equations. 5.9 Kinematics with Derivatives Motion in a line: displacement, velocity, acceleration, and total distance. 5.5, 5.10, 5.11 Integration & Area Anti-differentiation, definite integrals, and area enclosed between curves.