Figure 2.3: Mapping $y = f(x)$ to Reciprocal $y = \frac{1}{f(x)}$
1. The Reciprocal Transformation: $y = \frac{1}{f(x)}$
In IB DP Mathematics AA HL, sketching $y = \frac{1}{f(x)}$ given the visual curve or equation of $y = f(x)$ is tested heavily on Paper 1 (non-calculator). The geometric mapping rules must be memorized and applied systematically:
| Feature on Original $y = f(x)$ | Transformed Feature on $y = \frac{1}{f(x)}$ | Underlying Mathematical Justification |
|---|---|---|
| Zeros / $x$-intercepts: $(c, 0)$ | Vertical Asymptotes: $x = c$ | $\lim_{x \to c} \frac{1}{f(x)} = \pm\infty$. Signs on either side match the sign of $f(x)$! |
| Vertical Asymptotes: $x = k$ | $x$-intercepts: $(k, 0)$ (or open holes) | As $x \to k$, $|f(x)| \to \infty \implies \frac{1}{f(x)} \to 0$. |
| Invariant Points: $f(x) = 1$ or $f(x) = -1$ | Points of Intersection: $y = 1$ or $y = -1$ | $\frac{1}{1} = 1$ and $\frac{1}{-1} = -1$. The curves intersect exactly at these two horizontal lines. |
| Local Maximum: $(p, q)$ with $q > 0$ | Local Minimum: $\left(p, \frac{1}{q}\right)$ | $\frac{d}{dx}\left[\frac{1}{f(x)}\right] = -\frac{f'(x)}{[f(x)]^2} = 0 \iff f'(x) = 0$. Sign of slope inverts! |
| Local Minimum: $(u, v)$ with $v < 0$ | Local Maximum: $\left(u, \frac{1}{v}\right)$ | Reciprocal flips turning point nature while preserving the $x$-coordinate. |
| Horizontal Asymptote: $y = L$ ($L \neq 0$) | Horizontal Asymptote: $y = \frac{1}{L}$ | If $f(x) \to \pm\infty$ as $x \to \pm\infty$, then $\frac{1}{f(x)} \to 0$, giving horizontal asymptote $y = 0$. |
2. The Squared Transformation: $y = [f(x)]^2$
The squared mapping $y = [f(x)]^2$ creates distinct geometric and calculus features:
- Strict Non-Negativity: The entire curve satisfies $y \ge 0$ for all $x \in \text{dom}(f)$. No portion of the graph exists below the $x$-axis.
- Zeros Become Stationary Tangencies: By the chain rule, $\frac{d}{dx}[f(x)]^2 = 2f(x)f'(x)$. Whenever $f(x) = 0$, the derivative is strictly zero! Thus, every simple root of $f(x)$ becomes a local minimum and tangent to the $x$-axis.
- Invariant Points: Where $f(x) = 0 \implies y = 0$; where $f(x) = 1 \implies y = 1$. Any points where $f(x) = -1$ reflect to $y = (-1)^2 = 1$.
- Turning Points: If $(a, b)$ is a turning point on $f(x)$, then $(a, b^2)$ is a turning point on $[f(x)]^2$. Notice that if $b < 0$, a local minimum of $f(x)$ with negative height reflects and squares into a local maximum on $[f(x)]^2$!
3. Composite Transformations & Strict Order of Operations
A general composite transformation is represented as:
$$y = A \cdot f(B(x - C)) + D$$Horizontal vs. Vertical Execution
• Horizontal Transformations (Inner operations on $x$): Affect only $x$-coordinates. Always factor the coefficient of $x$: $f(Bx - E) = f\left(B\left(x - \frac{E}{B}\right)\right)$.
1. Horizontal stretch by scale factor $\frac{1}{|B|}$ (and reflection across $y$-axis if $B < 0$).
2. Horizontal translation by vector $\begin{pmatrix} C \\ 0 \end{pmatrix} = \begin{pmatrix} E/B \\ 0 \end{pmatrix}$.
• Vertical Transformations (Outer operations on $y$): Follow standard arithmetic order of operations (BODMAS):
1. Vertical stretch by scale factor $|A|$ (and reflection across $x$-axis if $A < 0$).
2. Vertical translation by vector $\begin{pmatrix} 0 \\ D \end{pmatrix}$.
4. Odd and Even Functions (Syllabus AHL 2.13)
Algebraic and graphical symmetry is tested both independently and within calculus integration questions:
- Even Functions: Satisfy $f(-x) = f(x)$ for all $x \in \text{dom}(f)$.
• Graphical Symmetry: Reflectional symmetry about the vertical $y$-axis ($x = 0$).
• Calculus Properties: The derivative of an even function is an odd function: $f'(-x) = -f'(x)$.
• Definite Integral: $\int_{-a}^a f(x)\,dx = 2\int_0^a f(x)\,dx$. - Odd Functions: Satisfy $f(-x) = -f(x)$ for all $x \in \text{dom}(f)$.
• Graphical Symmetry: Rotational symmetry of order 2 (180° rotation) about the origin $(0, 0)$.
• Origin Property: If $0 \in \text{dom}(f)$, then $f(0) = -f(0) \implies 2f(0) = 0 \implies \mathbf{f(0) = 0}$. (An odd function defined at $x = 0$ must pass through the origin).
• Calculus Properties: The derivative of an odd function is an even function: $f'(-x) = f'(x)$.
• Definite Integral: $\int_{-a}^a f(x)\,dx = 0$ (net signed area cancels). - Decomposition Theorem: Any function $f(x)$ with symmetric domain can be uniquely decomposed into an even part $f_E(x)$ and an odd part $f_O(x)$: $$f(x) = \underbrace{\frac{f(x) + f(-x)}{2}}_{f_E(x) \text{ (even)}} + \underbrace{\frac{f(x) - f(-x)}{2}}_{f_O(x) \text{ (odd)}}$$
Worked Examples
Worked Example 1: Reciprocal of a Cubic Polynomial (Paper 1 Style · 7 marks)
The function $f$ is a cubic polynomial with real coefficients. The graph of $y = f(x)$ has local maximum at $(-2, 4)$, local minimum at $(2, -1)$, and zeros at $x = -3$, $x = 0$, and $x = 3$.
(a) State the equations of all asymptotes of the graph of $g(x) = \frac{1}{f(x)}$. [3 marks]
(b) State the coordinates and nature of all local extrema of $g(x)$. [2 marks]
(c) Sketch the graph of $y = g(x)$, clearly showing all asymptotes and stationary points. [2 marks]
Official Markscheme Solution:
(a) Asymptotes: [3 marks]
• Vertical asymptotes occur where $f(x) = 0$: $x = -3$, $x = 0$, $x = 3$ [M1][A1].
• Horizontal asymptote: As $|x| \to \infty$, $|f(x)| \to \infty \implies \lim_{|x| \to \infty} \frac{1}{f(x)} = 0 \implies y = 0$ [A1].
Examiner Tip: Always write full equations with variables ($x = -3, x = 0, x = 3, y = 0$). Stating just numbers loses accuracy marks.
(b) Extrema: [2 marks]
• Local maximum on $f$ at $(-2, 4)$ maps to a local minimum at $\left(-2, \frac{1}{4}\right)$ [A1].
• Local minimum on $f$ at $(2, -1)$ maps to a local maximum at $\left(2, \frac{1}{-1}\right) = (2, -1)$ [A1].
(c) Sketch: [2 marks]
• 4 branches separated by the 3 vertical asymptotes $x = -3, 0, 3$ with correct asymptotic signs matching $f(x)$ [A1].
• Correct local minimum at $(-2, 0.25)$ and local maximum at $(2, -1)$, curve approaching $y = 0$ as $x \to \pm\infty$ [A1].
Worked Example 2: Squared Curve Analysis & Tangencies (Paper 1 Style · 6 marks)
Let $f(x) = x(x - 4)^2$. Consider the function $h(x) = [f(x)]^2$.
(a) Find the roots of $h(x) = 0$ and state their multiplicity. [2 marks]
(b) Using the derivative $h'(x)$, prove that the graph of $y = h(x)$ has horizontal tangents at all its $x$-intercepts. [2 marks]
(c) Determine the coordinates of all invariant points where $f(x) = h(x)$. [2 marks]
Official Markscheme Solution:
(a) Roots: [2 marks]
$h(x) = [x(x - 4)^2]^2 = x^2 (x - 4)^4$ [M1].
Roots are $x = 0$ (multiplicity 2) and $x = 4$ (multiplicity 4) [A1].
(b) Derivative Proof: [2 marks]
By chain rule: $h'(x) = 2 f(x) f'(x)$ [M1].
At any $x$-intercept, $f(x) = 0$.
Therefore, $h'(x) = 2(0) f'(x) = 0$ [R1].
Since the derivative is zero at $x = 0$ and $x = 4$, the tangent to $y = h(x)$ is horizontal ($y' = 0$) at every $x$-intercept [AG].
(c) Invariant Points: [2 marks]
$f(x) = [f(x)]^2 \iff f(x)[1 - f(x)] = 0 \iff f(x) = 0$ or $f(x) = 1$ [M1].
• $f(x) = 0 \implies x = 0$ and $x = 4$, yielding invariant points $(0, 0)$ and $(4, 0)$ [A1].
• $f(x) = 1 \implies x(x - 4)^2 = 1$. These points have coordinates $(x_i, 1)$ where $x_i$ are the real solutions to $x(x - 4)^2 = 1$.
Worked Example 3: Composite Transformation Sequence (Paper 1 Style · 5 marks)
The graph of $y = f(x)$ is transformed to the graph of $y = -3 f(2x + 8) - 5$.
(a) Describe in full the precise sequence of geometric transformations required. [4 marks]
(b) If $f(x)$ has a local maximum at $(1, 3)$, find the coordinates of the corresponding point on the transformed curve. [1 mark]
Official Markscheme Solution:
(a) Sequence of Transformations: [4 marks]
First factor the argument: $-3 f(2(x + 4)) - 5$ [M1].
1. Horizontal stretch by a scale factor of $\frac{1}{2}$ parallel to the $x$-axis [A1].
2. Horizontal translation by the vector $\begin{pmatrix} -4 \\ 0 \end{pmatrix}$ (or 4 units in the negative $x$-direction) [A1].
3. Vertical stretch by a scale factor of 3 parallel to the $y$-axis and reflection in the $x$-axis [A1].
4. Vertical translation by the vector $\begin{pmatrix} 0 \\ -5 \end{pmatrix}$ (or 5 units downwards) [A1].
(b) Transformed Point: [1 mark]
• New $x$-coordinate: $x_{\text{new}} = \frac{1}{2}(1) - 4 = 0.5 - 4 = -3.5$ [M1].
• New $y$-coordinate: $y_{\text{new}} = -3(3) - 5 = -9 - 5 = -14$.
Transformed point is $(-3.5, -14)$ [A1].
Worked Example 4: Parity Decomposition & Definite Integral (Paper 2 Style · 7 marks)
Let $f(x) = \frac{\sin x + x^2}{1 + x^2}$ for $x \in \mathbb{R}$.
(a) Express $f(x)$ as the sum of an even function $f_E(x)$ and an odd function $f_O(x)$. [3 marks]
(b) Hence evaluate the exact value of $\int_{-2}^2 f(x)\,dx$. [4 marks]
Official Markscheme Solution:
(a) Parity Decomposition: [3 marks]
$f(-x) = \frac{\sin(-x) + (-x)^2}{1 + (-x)^2} = \frac{-\sin x + x^2}{1 + x^2}$ [M1].
• $f_E(x) = \frac{f(x) + f(-x)}{2} = \frac{\frac{\sin x + x^2}{1 + x^2} + \frac{-\sin x + x^2}{1 + x^2}}{2} = \frac{2x^2}{2(1 + x^2)} = \frac{x^2}{1 + x^2}$ [A1].
• $f_O(x) = \frac{f(x) - f(-x)}{2} = \frac{\frac{\sin x + x^2}{1 + x^2} - \frac{-\sin x + x^2}{1 + x^2}}{2} = \frac{2\sin x}{2(1 + x^2)} = \frac{\sin x}{1 + x^2}$ [A1].
(b) Integral Evaluation: [4 marks]
$\int_{-2}^2 f(x)\,dx = \int_{-2}^2 f_E(x)\,dx + \int_{-2}^2 f_O(x)\,dx$ [M1].
Since $f_O(x)$ is continuous and odd on $[-2, 2]$, $\int_{-2}^2 f_O(x)\,dx = 0$ [R1].
Since $f_E(x)$ is even, $\int_{-2}^2 f_E(x)\,dx = 2 \int_0^2 \frac{x^2}{1 + x^2}\,dx$ [M1].
Use algebraic division: $\frac{x^2}{1 + x^2} = \frac{x^2 + 1 - 1}{x^2 + 1} = 1 - \frac{1}{1 + x^2}$ [M1].
$= 2 \left[ x - \arctan x \right]_0^2 = 2 (2 - \arctan 2 - 0) = 4 - 2\arctan 2$ [A1].
Worked Example 5: Rational Function Symmetries & Reciprocal Analysis (Paper 2 Style · 8 marks)
Let $f(x) = \frac{x}{x^2 - 1}$ for $x \in \mathbb{R} \setminus \{-1, 1\}$.
(a) Show algebraically that $f$ is an odd function and state its geometric symmetry. [2 marks]
(b) Determine the equations of all asymptotes of $y = f(x)$. [2 marks]
(c) Deduce the function $g(x) = \frac{1}{f(x)}$ and find all real solutions to $g(x) = x$. [4 marks]
Official Markscheme Solution:
(a) Odd Symmetry: [2 marks]
$f(-x) = \frac{-x}{(-x)^2 - 1} = \frac{-x}{x^2 - 1} = -f(x)$ [M1].
Since $f(-x) = -f(x)$ for all $x$ in its symmetric domain, $f$ is an odd function.
Geometric symmetry: Rotational symmetry of order 2 (180° rotational symmetry) about the origin $(0, 0)$ [A1].
(b) Asymptotes: [2 marks]
• Vertical asymptotes: $x^2 - 1 = 0 \implies x = 1$ and $x = -1$ [A1].
• Horizontal asymptote: $\lim_{x \to \pm\infty} \frac{x}{x^2 - 1} = 0 \implies y = 0$ [A1].
(c) Reciprocal & Equation Solving: [4 marks]
$g(x) = \frac{1}{f(x)} = \frac{x^2 - 1}{x} = x - \frac{1}{x}$ for $x \neq 0, \pm 1$ [A1].
Set $g(x) = x \implies x - \frac{1}{x} = x$ [M1].
$-\frac{1}{x} = 0$, which has no real solution [A1].
Alternatively: $x^2 - 1 = x^2 \implies -1 = 0$, a contradiction [R1].
Hence, there are no real solutions to $g(x) = x$.
Practice Exercises (10 Tiered Questions)
Work through these 10 examination-calibrated exercises ranging from foundational skills to challenging HL Paper 1 and Paper 2 problems. Expand each exercise to view the full IB markscheme and examiner advice.
The linear function $f(x) = 2x - 6$ has $x$-intercept 3 and $y$-intercept -6. State all asymptotes and the $y$-intercept of $g(x) = \frac{1}{f(x)}$.
Show Full Markscheme
• Zero of $f(x)$ occurs at $x = 3 \implies$ Vertical asymptote: $x = 3$ [A1].
• As $|x| \to \infty$, $f(x) \to \pm\infty \implies \frac{1}{f(x)} \to 0 \implies$ Horizontal asymptote: $y = 0$ [A1].
• $y$-intercept: $g(0) = \frac{1}{f(0)} = \frac{1}{-6} = -\frac{1}{6}$ [A1].
Examiner Advice: Asymptotes must always be stated as complete equations ($x = 3, y = 0$), not just values.
Determine algebraically whether $f(x) = x^4 - 3x^2 + 7$ is odd, even, or neither.
Show Full Markscheme
Substitute $-x$ into the function: [M1]
$f(-x) = (-x)^4 - 3(-x)^2 + 7 = x^4 - 3x^2 + 7$ [A1].
Since $f(-x) = f(x)$ for all $x \in \mathbb{R}$, $f$ is an even function [A1].
Examiner Advice: State the concluding identity $f(-x) = f(x)$ clearly to earn the final reasoning mark.
A curve $y = f(x)$ passes through $(-2, 0)$, $(1, 1)$, $(3, -1)$, and $(5, 2)$. State the coordinates of the corresponding points on $y = [f(x)]^2$.
Show Full Markscheme
Apply the transformation $(x, y) \to (x, y^2)$ to each coordinate: [M1]
• $(-2, 0) \to (-2, 0^2) = (-2, 0)$ [A1].
• $(1, 1) \to (1, 1^2) = (1, 1)$ [A1].
• $(3, -1) \to (3, (-1)^2) = (3, 1)$ [A1].
• $(5, 2) \to (5, 2^2) = (5, 4)$ [A1].
Examiner Advice: Note that $(3, -1)$ reflects across the $x$-axis because squaring converts negative ordinates to positive.
The graph of $f(x)$ has a local minimum at $(4, -2)$. State the coordinates and nature of the corresponding stationary point on $y = \frac{1}{f(x)}$.
Show Full Markscheme
• The reciprocal transformation inverts the $y$-coordinate while preserving the $x$-coordinate: $y = \frac{1}{-2} = -0.5$ [M1].
• Nature of stationary point: Since $\frac{d}{dx}\left[\frac{1}{f(x)}\right] = -\frac{f'(x)}{[f(x)]^2}$, the sign of the slope flips on either side of the point. A local minimum with negative height becomes a local maximum at $(4, -0.5)$ [A1].
Examiner Advice: A common trap is assuming that because $y$ is negative, the nature doesn't flip. All local extrema flip nature under the reciprocal mapping!
Let $f(x)$ be an odd function and $g(x)$ be an even function. Prove algebraically that the product $h(x) = f(x)g(x)$ is an odd function.
Show Full Markscheme
Evaluate $h(-x)$: [M1]
$h(-x) = f(-x) \cdot g(-x)$ [M1].
Since $f$ is odd, $f(-x) = -f(x)$. Since $g$ is even, $g(-x) = g(x)$ [R1].
$h(-x) = (-f(x)) \cdot g(x) = -(f(x)g(x)) = -h(x)$ [A1].
Since $h(-x) = -h(x)$, $h(x)$ is an odd function [AG].
Let $g(x) = [f(x)]^2$. Use the chain rule to show that $g'(x) = 0$ whenever $f(x) = 0$, and explain the graphical significance.
Show Full Markscheme
By the chain rule: $g'(x) = 2f(x)f'(x)$ [M1].
When $f(x) = 0$: $g'(x) = 2(0)f'(x) = 0$ (assuming $f'(x)$ exists) [A1].
Graphical significance: Every real root of $f(x)$ becomes a stationary turning point on $y = [f(x)]^2$. Because $[f(x)]^2 \ge 0$, the curve touches and is tangent to the $x$-axis, forming a local minimum at each zero without crossing below the axis [R1].
Given $f(x) = 2\sin x + 1$ for $0 \le x \le 2\pi$, find the equations of all vertical asymptotes of $y = \frac{1}{f(x)}$.
Show Full Markscheme
Vertical asymptotes occur where $f(x) = 0$: [M1]
$2\sin x + 1 = 0 \implies \sin x = -\frac{1}{2}$ [M1].
In the domain $[0, 2\pi]$, the reference angle is $\arcsin(1/2) = \frac{\pi}{6}$.
Sine is negative in Quadrants 3 and 4:
• Quadrant 3: $x = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$ [A1].
• Quadrant 4: $x = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$ [A1].
Vertical asymptotes: $x = \frac{7\pi}{6}$ and $x = \frac{11\pi}{6}$.
Let $f$ and $g$ be two functions. Prove that:
(a) If $g$ is an even function, then $(f \circ g)$ is an even function for any function $f$.
(b) If both $f$ and $g$ are odd functions, then $(f \circ g)$ is an odd function.
Show Full Markscheme
(a) Even inner function: [2 marks]
Consider $(f \circ g)(-x) = f(g(-x))$ [M1].
Since $g$ is even, $g(-x) = g(x)$.
Therefore, $f(g(-x)) = f(g(x)) = (f \circ g)(x)$ [A1].
Hence $(f \circ g)$ is an even function [AG].
(b) Both functions odd: [2 marks]
Consider $(f \circ g)(-x) = f(g(-x))$ [M1].
Since $g$ is odd, $g(-x) = -g(x)$.
$= f(-g(x))$. Since $f$ is odd, $f(-u) = -f(u)$.
$= -f(g(x)) = -(f \circ g)(x)$ [A1].
Hence $(f \circ g)$ is an odd function [AG].
The function $f(x) = \frac{x^2 - 1}{x + 2}$ has an oblique asymptote. Describe the asymptotic behavior of $g(x) = \frac{1}{f(x)}$ as $x \to \pm\infty$, and find its horizontal asymptote.
Show Full Markscheme
Construct the reciprocal function: [M1]
$g(x) = \frac{1}{f(x)} = \frac{x + 2}{x^2 - 1}$ for $x \neq \pm 1$ [A1].
In $g(x)$, the degree of the numerator (1) is strictly less than the degree of the denominator (2) [R1].
Take the limits as $x \to \pm\infty$:
$\lim_{x \to \pm\infty} \frac{x + 2}{x^2 - 1} = \lim_{x \to \pm\infty} \frac{\frac{1}{x} + \frac{2}{x^2}}{1 - \frac{1}{x^2}} = \frac{0}{1} = 0$ [M1].
Horizontal asymptote: $y = 0$ [A1].
Key Takeaway: Whenever $f(x)$ has an oblique asymptote ($|f(x)| \to \infty$ as $|x| \to \infty$), its reciprocal $y = \frac{1}{f(x)}$ always possesses the horizontal asymptote $y = 0$.
Use odd and even parity properties to evaluate the exact value of:
(a) $\int_{-\pi/4}^{\pi/4} x^5 \cos(x^2)\,dx$
(b) $\int_{-2}^2 (3x^2 + 5x^3)\,dx$
Show Full Markscheme
(a) Odd integrand: [2 marks]
Let $k(x) = x^5 \cos(x^2)$. Test parity: [M1]
$k(-x) = (-x)^5 \cos((-x)^2) = -x^5 \cos(x^2) = -k(x)$.
Since $k(x)$ is continuous and odd, its integral over any symmetric interval $[-a, a]$ is strictly zero:
$\int_{-\pi/4}^{\pi/4} x^5 \cos(x^2)\,dx = 0$ [A1].
(b) Splitting into even and odd components: [3 marks]
$\int_{-2}^2 (3x^2 + 5x^3)\,dx = \int_{-2}^2 3x^2\,dx + \int_{-2}^2 5x^3\,dx$ [M1].
• $5x^3$ is an odd function, so $\int_{-2}^2 5x^3\,dx = 0$ [R1].
• $3x^2$ is an even function, so $\int_{-2}^2 3x^2\,dx = 2 \int_0^2 3x^2\,dx = 6 \left[\frac{x^3}{3}\right]_0^2 = 2(2^3) = 16$ [A1].
Total value: $16 + 0 = 16$ [A1].
IB Formula Booklet References (Topic 2)
Syllabus 2.13: Function Transformations
The official IB DP Mathematics AA Formula Booklet provides basic translation and stretch notation:
- $y = f(x) + d$: Vertical translation by $\begin{pmatrix} 0 \\ d \end{pmatrix}$.
- $y = f(x - c)$: Horizontal translation by $\begin{pmatrix} c \\ 0 \end{pmatrix}$.
- $y = a f(x)$: Vertical stretch by scale factor $a$.
- $y = f(b x)$: Horizontal stretch by scale factor $\frac{1}{b}$.
Crucial Note: The Formula Booklet does NOT include rules for reciprocal curves $y = \frac{1}{f(x)}$, squared curves $y = [f(x)]^2$, or algebraic tests for odd/even functions. These must be committed to memory.
Key Formulas & Identities
1. Reciprocal Transformation Checklist: $y = \frac{1}{f(x)}$
- $f(x) = 0 \implies$ Vertical asymptote $x = c$.
- Vertical asymptote on $f(x) \implies x$-intercept on $\frac{1}{f(x)}$.
- Invariant points: $f(x) = 1 \implies y = 1$; $f(x) = -1 \implies y = -1$.
- Local maximum $(p, q)$ with $q > 0 \implies$ Local minimum $\left(p, \frac{1}{q}\right)$.
- Local minimum $(u, v)$ with $v < 0 \implies$ Local maximum $\left(u, \frac{1}{v}\right)$.
- Sign rule: Sign of $\frac{1}{f(x)}$ is identical to the sign of $f(x)$ across all intervals.
2. Squared Transformation Checklist: $y = [f(x)]^2$
- Range is non-negative: $y \ge 0$ everywhere.
- Zeros become stationary tangencies: $\frac{d}{dx}[f(x)]^2 = 2f(x)f'(x) = 0$ at all roots.
- Invariant points: $(x, 0) \to (x, 0)$ and $(x, 1) \to (x, 1)$. Also $(x, -1) \to (x, 1)$.
- Turning point $(a, b) \to (a, b^2)$. If $b < 0$, minimum flips to maximum!
3. Odd and Even Function Symmetry Checklist
- Even Function: $f(-x) = f(x)$ $\implies$ Reflectional symmetry across the $y$-axis ($x = 0$). Integral: $\int_{-a}^a f(x)\,dx = 2\int_0^a f(x)\,dx$.
- Odd Function: $f(-x) = -f(x)$ $\implies$ 180° rotational symmetry about origin $(0, 0)$. If $f(0)$ exists, $f(0) = 0$. Integral: $\int_{-a}^a f(x)\,dx = 0$.
- Calculus Parity Inversion: $\frac{d}{dx}[\text{even}] = \text{odd}$; $\frac{d}{dx}[\text{odd}] = \text{even}$.
- Decomposition: $f_E(x) = \frac{f(x) + f(-x)}{2}$, $f_O(x) = \frac{f(x) - f(-x)}{2}$.
Common Exam Pitfalls
Pitfall 1: Sign Discrepancies on Reciprocal Curves
Remember that $\frac{1}{\text{negative}} = \text{negative}$. When $f(x)$ is negative, $\frac{1}{f(x)}$ must also be negative. Students frequently reflect the reciprocal curve into the upper half-plane by mistake!
Pitfall 2: Value of Odd Function at the Origin
If an odd function is defined at $x = 0$, its value must be 0: $f(0) = -f(0) \implies 2f(0) = 0 \implies f(0) = 0$. If a given function has $f(0) \neq 0$ (for instance, $f(0) = 3$), it can never be an odd function!
Pitfall 3: Tangency vs. Sharp Corner on $[f(x)]^2$
Do not sketch squared graphs with sharp V-shaped corners at the $x$-axis like an absolute value graph $|f(x)|$. Because $\frac{d}{dx}[f(x)]^2 = 2f(x)f'(x) = 0$ at roots, the curve must be smoothly parabolic and horizontal at every zero!
Topic Diagnostic Test
Complete this timed diagnostic test to verify your conceptual command of reciprocal graphs, squared mappings, and function symmetry.
Section A: Diagnostic Multiple Choice Questions
If $f(x)$ has a local minimum at $(2, 4)$, then $y = \frac{1}{f(x)}$ has:
Which of the following functions is neither even nor odd?
If the graph of $y = f(x)$ has $x$-intercepts at $x = -1$ and $x = 3$, what can be deduced about the graph of $y = [f(x)]^2$ at these points?
Let $f$ be an even function and $g$ be an odd function. What is the parity of the composite function $h(x) = f(g(x))$?
Section B: Examination-Style Free Response Questions
The function $f$ is defined by $f(x) = x\sqrt{4 - x^2}$ for $-2 \le x \le 2$.
(a) Show algebraically that $f$ is an odd function. [2 marks]
(b) Find the stationary points of $f(x)$ and determine their nature. [3 marks]
(c) Hence sketch the graph of $f(x)$ showing all key features. [1 mark]
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(a) Odd function proof: [2 marks]
$f(-x) = (-x)\sqrt{4 - (-x)^2} = -x\sqrt{4 - x^2} = -f(x)$ [M1].
Since $f(-x) = -f(x)$ for all $x \in [-2, 2]$, $f$ is an odd function [A1].
(b) Stationary points: [3 marks]
$f'(x) = 1\sqrt{4 - x^2} + x\left(\frac{-2x}{2\sqrt{4 - x^2}}\right) = \frac{4 - x^2 - x^2}{\sqrt{4 - x^2}} = \frac{4 - 2x^2}{\sqrt{4 - x^2}}$ [M1].
$f'(x) = 0 \implies 4 - 2x^2 = 0 \implies x^2 = 2 \implies x = \pm\sqrt{2}$ [A1].
• At $x = \sqrt{2}$: $y = \sqrt{2}\sqrt{4 - 2} = 2$. Local maximum at $(\sqrt{2}, 2)$.
• At $x = -\sqrt{2}$: by odd symmetry, local minimum at $(-\sqrt{2}, -2)$ [A1].
(c) Sketch: [1 mark]
Curve passing through endpoints $(-2, 0)$ and $(2, 0)$, passing through $(0, 0)$ with 180° rotational symmetry [A1].
Let $f(x) = \frac{1}{x^2 - 4}$.
(a) State the equations of all asymptotes of $y = f(x)$. [2 marks]
(b) Sketch $y = f(x)$ and on the same axes sketch $y = [f(x)]^2$. [3 marks]
(c) Find the coordinates of all invariant points where $f(x) = [f(x)]^2$. [3 marks]
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(a) Asymptotes: [2 marks]
Vertical asymptotes: $x^2 - 4 = 0 \implies x = 2$ and $x = -2$ [A1].
Horizontal asymptote: $\lim_{x \to \pm\infty} \frac{1}{x^2 - 4} = 0 \implies y = 0$ [A1].
(b) Sketch comparison: [3 marks]
• $f(x)$ has vertical asymptotes at $x = \pm 2$, local max at $(0, -0.25)$, branches above $y = 0$ for $|x| > 2$ [A1].
• $[f(x)]^2 = \frac{1}{(x^2 - 4)^2}$ is strictly positive everywhere, asymptotes at $x = \pm 2, y = 0$, local min at $(0, 0.0625)$ [A1][A1].
(c) Invariant points: [3 marks]
$f(x) = [f(x)]^2 \iff f(x)[1 - f(x)] = 0$ [M1].
Since $f(x) = \frac{1}{x^2 - 4} \neq 0$ for all real $x$, we require $f(x) = 1$ [M1].
$\frac{1}{x^2 - 4} = 1 \implies x^2 - 4 = 1 \implies x^2 = 5 \implies x = \pm\sqrt{5}$ [A1].
Invariant points: $(\sqrt{5}, 1)$ and $(-\sqrt{5}, 1)$.