What is build the reciprocal graph, stage by stage?

The reciprocal y = (1)/(f) is the most-examined combination, and the safest way to sketch it is to mark the structure from f first, then draw. Below, y = (1)/(x^2 - 1) is built from f(x) = x^2 - 1 one decision at a time. (The 2022 HSC asked exactly this kind of "describe the features of (1)/(f)" question.)

What are square has zeros, not just minimums?

f^2 has zeros wherever f does. The graph touches the x-axis at each one rather than crossing.

NESA 2021 HSC Q15-style practice: Given f(x) = x and g(x) = sin x, sketch y = f(x) g(x) = x sin x for -2π x 2π.

Zeros of x sin x occur where x = 0 or sin x = 0, that is at x = 0, π, 2 π. The amplitude grows linearly: |x sin x| |x|, with equality at sin x = 1. The envelope is y = x, so the graph oscillates between y = -|x| and y = |x|, touching those lines at x = (π)/(2) + k π. Markers expect the zeros marked, the envelope drawn, and the oscillation shown with growing amplitude.

§-Syllabus dot point

NSWMaths AdvancedSyllabus dot point

How do we sketch graphs built from sums, differences, products, quotients and reciprocals of standard functions?

Sketch graphs of sums, differences, products, quotients, squares and reciprocals of two known functions

A focused answer to the HSC Maths Advanced dot point on combining functions graphically. How to build sketches of $f + g$ , $f g$ , $f / g$ , $1 / f$ and $f^2$ from the graphs of $f$ and $g$ , where features come from, and what asymptotes and zeros do, with worked examples.

Generated by Claude Opus 4.814 min answerUpdated 2026-06-20

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

NESA wants you to sketch a function built by combining two known functions. The combinations are sums $f + g$ , differences $f - g$ , products $f g$ , quotients $\frac{f}{g}$ , squares $f^2$ , and reciprocals $\frac{1}{f}$ . You need to read features off the parent graphs and predict the combined graph.

The answer

Sums and differences

For $y = f(x) + g(x)$ , add the heights of the two graphs at each $x$ . Useful checks:

Zeros of $f + g$ occur where the two graphs are reflections of each other through the $x$ -axis, that is $f(x) = -g(x)$ , not generally where either function is zero alone.
If $f$ is bounded and $g$ is unbounded, the long-term behaviour of $f + g$ is the same as $g$ .
For $f - g$ , subtract the heights. At points where $f = g$ , the difference is zero.

Products

For $y = f(x) g(x)$ , multiply the heights:

Zeros of $f g$ are the union of the zeros of $f$ and $g$ .
The sign of $f g$ follows the rule of signs: $(+)(+) = +$ , $(+)(-) = -$ , and so on.
If one factor is bounded between $-1$ and $1$ (like $\sin x$ ), the other factor acts as an envelope: $|f g| \le |f|$ and the graph oscillates between $y = \pm f(x)$ .
If both factors grow, $f g$ grows faster.

Quotients

For $y = \frac{f(x)}{g(x)}$ :

Zeros of $\frac{f}{g}$ are the zeros of $f$ , provided $g$ is non-zero there.
Vertical asymptotes occur at zeros of $g$ where $f$ is non-zero.
If $g(x) = 0$ and $f(x) = 0$ at the same point, there is a hole or a finite limit; check carefully.
Horizontal asymptotes come from the ratio of long-term behaviours: $\frac{f}{g} \to 0$ if $g$ grows faster, a non-zero constant if they grow at the same rate, and $\pm \infty$ if $f$ grows faster.

Reciprocals

The graph of $y = \frac{1}{f(x)}$ comes from $y = f(x)$ by these rules:

Where $f(x) = 0$ , $\frac{1}{f}$ has a vertical asymptote.
Where $f(x) = \pm 1$ , the reciprocal also equals $\pm 1$ , so the two graphs meet on the lines $y = 1$ and $y = -1$ .
$\frac{1}{f}$ has the same sign as $f$ everywhere $f \neq 0$ .
Local maxima of $f$ where $f > 0$ become local minima of $\frac{1}{f}$ , and vice versa, because reciprocating flips relative size.
As $f(x) \to \pm \infty$ , $\frac{1}{f(x)} \to 0$ . As $f(x) \to 0^{\pm}$ , $\frac{1}{f(x)} \to \pm \infty$ .

Key fact

To sketch a combined graph, read it off the parents feature by feature: zeros, signs, and turning points. The single most-tested case is the reciprocal $\frac{1}{f}$ , governed by four rules: a zero of $f$ becomes a vertical asymptote of $\frac{1}{f}$ ; where $f = \pm 1$ the two graphs meet (since $\frac{1}{\pm 1} = \pm 1$ ); $\frac{1}{f}$ keeps the same sign as $f$ ; and a local maximum of $f$ (where $f > 0$ ) becomes a local minimum of $\frac{1}{f}$ , because reciprocating turns big into small. For other combinations: $f + g$ adds heights, $fg$ multiplies them (a factor bounded by $1$ , like $\sin x$ , acts as an envelope $y = \pm f$ ), $\frac{f}{g}$ puts zeros of $f$ on the axis and zeros of $g$ at asymptotes, and $f^2$ folds the graph above the axis with double-root touches at the zeros of $f$ .

Squares

For $y = (f(x))^2$ :

Zeros of $f^2$ are the zeros of $f$ , but now they are double roots: the graph touches the $x$ -axis and turns.
$f^2 \ge 0$ everywhere, so the graph never dips below the $x$ -axis.
Where $f(x) = \pm 1$ , $f^2(x) = 1$ . Where $|f(x)| > 1$ , $f^2 > |f|$ . Where $|f(x)| < 1$ , $f^2 < |f|$ .
Extrema of $f^2$ occur where $f f' = 0$ , that is where $f = 0$ or where $f$ has an extremum.

Build the reciprocal graph, stage by stage

The reciprocal $y = \frac{1}{f}$ is the most-examined combination, and the safest way to sketch it is to mark the structure from $f$ first, then draw. Below, $y = \frac{1}{x^2 - 1}$ is built from $f(x) = x^2 - 1$ one decision at a time. (The 2022 HSC asked exactly this kind of "describe the features of $\frac{1}{f}$ " question.)

Stage 1, plot the parent function. Draw $y = f(x) = x^2 - 1$ , an upward parabola. Mark the two features that drive the reciprocal: the zeros at $x = -1$ and $x = 1$ , and the minimum at $(0, -1)$ . Everything about $\frac{1}{f}$ follows from these.

Stage 2, turn zeros into asymptotes and mark the meeting lines. Each zero of $f$ (at $x = \pm 1$ ) becomes a vertical asymptote of $\frac{1}{f}$ , because dividing by zero blows up. Draw those dashed verticals. Also draw the lines $y = 1$ and $y = -1$ : wherever $f = \pm 1$ , the reciprocal equals $\pm 1$ too, so $\frac{1}{f}$ and $f$ cross on these lines.

Stage 3, sketch the three branches. Now draw $\frac{1}{f}$ between and outside the asymptotes. Outside $(\,x < -1$ and $x > 1)$ the parabola is positive and large, so the reciprocal is positive and small, hugging the $x$ -axis far out and rising to $+\infty$ at each asymptote. Between the asymptotes $f$ is negative (down to $-1$ at the centre), so $\frac{1}{f}$ is negative, plunging to $-\infty$ near each asymptote.

Stage 4, check the extremum flips. The minimum of $f$ at $(0, -1)$ , where $f$ is negative, becomes a local maximum of $\frac{1}{f}$ at the same point $(0, -1)$ : $\frac{1}{-1} = -1$ , and reciprocating a "most negative" value gives the "least negative" one. This sign-aware flip of turning points is the detail markers look for.

How exam questions ask about combining functions

The combinations come up in a handful of standard phrasings:

"Describe the key features of $y = \frac{1}{f(x)}$ " given the zeros and sign of $f$ . Zeros become vertical asymptotes, the sign is preserved, the curves meet at $y = \pm 1$ , and turning points flip (max becomes min where $f > 0$ ). State each (2022 HSC Q14).
"Sketch $y = f(x)\,g(x)$ " with one factor a sine or cosine. Mark zeros (the union of both factors' zeros), draw the envelope $y = \pm f$ , and show the oscillation with growing or decaying amplitude (2021 HSC Q15).
"Sketch $y = f(x) + g(x)$ ." Add ordinates: at each $x$ , add the two heights. Crossings of the $x$ -axis occur where $f = -g$ , not where either is zero.
"Sketch $y = \frac{f(x)}{g(x)}$ ." Zeros of $f$ give $x$ -intercepts; zeros of $g$ give vertical asymptotes; a common zero may be a hole; compare growth rates for any horizontal asymptote.
"Sketch $y = [f(x)]^2$ ." Fold the graph above the $x$ -axis; the zeros of $f$ become double-root touches, and $y$ never goes negative.
"On the diagram of $y = f(x)$ , sketch $y = \frac{1}{f(x)}$ " (a build-on-the-given-graph task). Read the zeros, the $\pm 1$ crossings and the turning points straight off the printed curve and apply the four reciprocal rules.

Worked example

Six graphs built by combining simpler functions, each illustrating a different combining operation and its effect on key features, ending with a damped oscillation in an Australian context.

Sketch the reciprocal of $\sin x$

To draw $y = \csc x = \frac{1}{\sin x}$ , first identify what zeros of $\sin x$ do to $\frac{1}{\sin x}$ : division by zero creates vertical asymptotes. So vertical asymptotes occur at $x = k\pi$ for integer $k$ .

Where $\sin x = \pm 1$ (at $x = \frac{\pi}{2} + k\pi$ ), the reciprocal also equals $\pm 1$ , so the two graphs meet there. Between each pair of asymptotes, $\csc x$ has the same sign as $\sin x$ and curves away from the $x$ -axis rather than toward it.

Sketch the square of $\sin x$

For $y = \sin^2 x$ , squaring does two things: it makes every value non-negative (the graph lives above or on the $x$ -axis), and it doubles the original zeros into touching points rather than crossings.

Zeros of $\sin x$ at $x = k\pi$ become double-root touches of $\sin^2 x$ . Maxima of $\sin x$ at $\pm 1$ become maxima of $\sin^2 x$ at $1$ . Because each full cycle of $\sin x$ produces two identical half-arches in $\sin^2 x$ , the period halves from $2\pi$ to $\pi$ .

Analyse the quotient $y = \frac{\sin x}{x}$

Both numerator and denominator vanish at $x = 0$ , so there is a hole rather than an asymptote there. Using the standard limit, $\lim_{x \to 0} \frac{\sin x}{x} = 1$ , so the hole occurs at $(0, 1)$ .

For $x \neq 0$ , zeros come from $\sin x = 0$ , giving $x = \pm\pi, \pm 2\pi, \dots$ As $|x| \to \infty$ , $\frac{1}{x} \to 0$ , so the oscillation amplitude decays like $\frac{1}{|x|}$ and the graph flattens towards the $x$ -axis.

Sketch the product $y = e^{-x} \sin x$ for $x \ge 0$

The factor $\sin x$ oscillates between $-1$ and $1$ , so the product $e^{-x} \sin x$ oscillates between $-e^{-x}$ and $e^{-x}$ . These bounds are the envelopes: the curves $y = \pm e^{-x}$ form the outer boundary that the graph touches wherever $\sin x = \pm 1$ .

Zeros of the product occur wherever $\sin x = 0$ , namely at $x = k\pi$ . Between each pair of consecutive zeros the curve makes one arch, with decreasing peak height because $e^{-x}$ is decaying.

Sketch the sum $y = x + \sin x$

Adding $\sin x$ to the line $y = x$ shifts each point on the line up or down by at most $1$ . The straight line $y = x$ acts as a "spine" that the graph oscillates around, never straying further than $1$ unit away.

Crossings of the line $y = x$ occur where $\sin x = 0$ , that is at $x = k\pi$ . The graph is always increasing because $(x + \sin x)' = 1 + \cos x \ge 0$ for all $x$ .

Sketch a damped suspension oscillation (Australian context)

After a car hits a speed hump on a Melbourne street, the height of its body above rest can be modelled as the product $h(t) = 8 e^{-0.6 t} \cos(4 t)$ (in centimetres, $t$ in seconds): a cosine wobble multiplied by a decaying exponential. Sketch its shape using the product rules.

The bounded factor here is $\cos(4t)$ , which stays between $-1$ and $1$ . So the other factor, $8 e^{-0.6t}$ , acts as the envelope: the graph is squeezed between $y = 8 e^{-0.6t}$ and $y = -8 e^{-0.6t}$ , touching the upper envelope whenever $\cos(4t) = 1$ and the lower one whenever $\cos(4t) = -1$ .

The zeros of $h$ come from the bounded factor: $\cos(4t) = 0$ at $4t = \frac{\pi}{2} + k\pi$ , that is $t = \frac{\pi}{8} + \frac{k\pi}{4}$ , where the suspension passes through its rest height. Each oscillation has a smaller amplitude than the last because $e^{-0.6t}$ is decaying, so the wobble dies away: a settling suspension, exactly as the envelope $\pm 8 e^{-0.6t}$ predicts, starting at $h(0) = 8$ cm.

Key features summary: $\csc x$ has asymptotes at zeros of $\sin x$ ; $\sin^2 x$ has period $\pi$ and touching zeros; $\frac{\sin x}{x}$ has a hole at $(0,1)$ and decaying oscillations; $e^{-x}\sin x$ has exponential envelopes $y = \pm e^{-x}$ ; $x + \sin x$ oscillates within $1$ of the line $y = x$ ; the suspension model $8e^{-0.6t}\cos(4t)$ decays inside the envelope $y = \pm 8 e^{-0.6t}$ with zeros at $t = \frac{\pi}{8} + \frac{k\pi}{4}$ .

Common traps

Treating reciprocals like reflections: $\frac{1}{f}$ is not a reflection. The shape distorts: large values become small and vice versa.
Forgetting asymptotes are about zeros of the denominator: For $\frac{f}{g}$ , vertical asymptotes come from $g(x) = 0$ , not $f(x) = 0$ . Zeros come from $f(x) = 0$ (with $g(x) \neq 0$ ).
Square has zeros, not just minimums: $f^2$ has zeros wherever $f$ does. The graph touches the $x$ -axis at each one rather than crossing.
Misreading envelopes: For $y = f g$ with $|g| \le 1$ , the envelope is $y = \pm f$ , not $y = f$ alone. Both branches matter.
Missing holes versus asymptotes in quotients: If $f$ and $g$ both vanish at the same point with a common factor, you get a hole, not an asymptote. Factor and cancel before concluding.

Exam technique

These are "read it off the parents" questions, and the marks sit on specific transferred features.

List the features before you draw. For any combination, find the zeros, the sign on each interval, and the turning points of the result from the parents first; then the curve almost draws itself.
For a reciprocal, run the four-rule checklist. Zeros to asymptotes; same sign as $f$ ; meet $f$ at $y = \pm 1$ ; max becomes min where $f > 0$ (and the reverse where $f < 0$ ). State each, do not just draw.
For a product with $\sin$ or $\cos$ , draw the envelope as a guide. Faint $y = \pm f$ lines, then the oscillation touching them at the peaks. Mark the zeros from both factors.
For a quotient, separate zeros from asymptotes. $x$ -intercepts come from the numerator's zeros; vertical asymptotes from the denominator's zeros; a shared zero is a hole, so factor and cancel to check.
Use the line $y = \pm 1$ as anchor points. A reciprocal always passes through the same points as $f$ on those lines, giving you exact crossings to pin the sketch.
Keep $f^2 \ge 0$ . The square never dips below the axis; the original zeros become touching points, not crossings.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2022 HSC Q143 marksThe function

f

has zeros at

x = -2

and

x = 3

and is positive elsewhere. Describe the key features of the graph of

y = \frac{1}{f(x)}

Show worked answer →

Zeros of $f$ become vertical asymptotes of $\frac{1}{f}$ , so vertical asymptotes occur at $x = -2$ and $x = 3$ .

The sign of $\frac{1}{f}$ matches the sign of $f$ , so $\frac{1}{f}$ is positive on $(-\infty, -2)$ , $(3, \infty)$ , and wherever $f$ is positive in $(-2, 3)$ (the whole interior if the question's "positive elsewhere" includes that interval, otherwise piecewise).

Local maxima of $f$ become local minima of $\frac{1}{f}$ in regions where $f > 0$ , and vice versa, because reciprocating flips relative size.

Markers reward identifying asymptotes from zeros, sign matching, and the inversion of extrema.

2021 HSC Q153 marksGiven

f(x) = x

and

g(x) = \sin x

, sketch

y = f(x) g(x) = x \sin x

for

-2\pi \le x \le 2\pi

Show worked answer →

Zeros of $x \sin x$ occur where $x = 0$ or $\sin x = 0$ , that is at $x = 0, \pm \pi, \pm 2 \pi$ .

The amplitude grows linearly: $|x \sin x| \le |x|$ , with equality at $\sin x = \pm 1$ .

The envelope is $y = \pm x$ , so the graph oscillates between $y = -|x|$ and $y = |x|$ , touching those lines at $x = \frac{\pi}{2} + k \pi$ .

Markers expect the zeros marked, the envelope drawn, and the oscillation shown with growing amplitude.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marksThe function

f(x) = x - 2

has a single zero at

x = 2

. State the equation of the vertical asymptote of

y = \dfrac{1}{f(x)}

and give the sign of

\dfrac{1}{f(x)}

on each side of it.

Show worked solution →

A zero of $f$ becomes a vertical asymptote of the reciprocal. Since $f(x) = x - 2 = 0$ only at $x = 2$ , dividing by zero there blows the reciprocal up:

\text{vertical asymptote: } x = 2.

The reciprocal keeps the sign of $f$ . For $x < 2$ , $f(x) = x - 2 < 0$ , so $\dfrac{1}{f(x)} < 0$ ; for $x > 2$ , $f(x) > 0$ , so $\dfrac{1}{f(x)} > 0$ .

Marker's note: one mark for the asymptote $x = 2$ (from the zero of $f$ ), one for the sign on both sides matching the sign of $f$ . Stating an asymptote without the sign analysis earns only the first mark.

foundation3 marksLet

f(x) = x + 1

and

g(x) = x - 3

. (a) Write

y = f(x) + g(x)

in simplest form and state its zero. (b) State the zeros of

y = f(x)\,g(x)

Show worked solution →

Part (a): add the two functions.

f(x) + g(x) = (x + 1) + (x - 3) = 2x - 2.

The sum is zero when $2x - 2 = 0$ , that is at $x = 1$ . (Note this is where $f(x) = -g(x)$ , not where either parent is zero.)

Part (b): zeros of a product are the union of the parents' zeros. The product $f(x)\,g(x) = (x + 1)(x - 3)$ is zero when either factor is zero:

x = -1 \quad \text{or} \quad x = 3.

Marker's note: one mark for the simplified sum $2x - 2$ and its zero $x = 1$ , one for recognising product zeros as the union, one for both values $x = -1$ and $x = 3$ . Writing the sum's zero as $x = -1$ or $x = 3$ confuses the two operations and loses the mark.

foundation2 marksFor

f(x) = x - 4

, describe how the graph of

y = [f(x)]^2

behaves at

x = 4

and state why

y

is never negative.

Show worked solution →

Square the linear factor.

[f(x)]^2 = (x - 4)^2.

At the zero of $f$ , the square gives a double root. Because $(x - 4)^2 = 0$ has a repeated root at $x = 4$ , the graph touches the $x$ -axis at $(4, 0)$ and turns, rather than crossing.

The output is a square, so it is never negative. For every real $x$ , $(x - 4)^2 \ge 0$ , so the whole graph sits on or above the $x$ -axis.

Marker's note: one mark for the touching (double-root) behaviour at $x = 4$ , one for justifying $y \ge 0$ from the square. Describing a crossing rather than a touch loses the first mark.

core4 marksConsider

f(x) = x

and

g(x) = \cos x

over

0 \le x \le 2\pi

, and the product

y = f(x)\,g(x) = x\cos x

. (a) State the two envelope curves between which the graph oscillates. (b) Find all zeros of

y

in the interval. (c) State the exact

x

-values in the interval at which the graph touches an envelope.

Show worked solution →

Part (a): the bounded factor sets the envelope. Since $-1 \le \cos x \le 1$ , multiplying by $x$ (with $x \ge 0$ here) gives $-x \le x\cos x \le x$ , so the graph oscillates between the envelope curves

y = x \quad \text{and} \quad y = -x.

Part (b): zeros come from either factor. $x\cos x = 0$ when $x = 0$ or $\cos x = 0$ . On $0 \le x \le 2\pi$ , $\cos x = 0$ at $x = \dfrac{\pi}{2}$ and $x = \dfrac{3\pi}{2}$ , so the zeros are

x = 0, \quad x = \frac{\pi}{2}, \quad x = \frac{3\pi}{2}.

Part (c): touches occur where $\cos x = \pm 1$ . At those points $x\cos x = \pm x$ , meeting an envelope. On the interval $\cos x = 1$ at $x = 0$ and $x = 2\pi$ , and $\cos x = -1$ at $x = \pi$ :

x = 0, \quad x = \pi, \quad x = 2\pi.

Marker's note: one mark for the envelopes $y = \pm x$ , two for the correct zeros in (b), one for the touch points in (c). Working in degrees, or dropping $x = 2\pi$ at the endpoint, is the usual slip.

core4 marksThe table below gives values of two functions

f

and

g

at five points.

x = -2

f = 3

g = -1

x = -1

f = 0

g = 1

x = 0

f = -1

g = 2

x = 1

f = 0

g = 1

x = 2

f = 3

g = -1

. (a) Complete a table of values for

y = \dfrac{f(x)}{g(x)}

at these five points. (b) State any

x

-value from the table where

y = \dfrac{f(x)}{g(x)}

has an

x

-intercept, and explain why

g

produces no vertical asymptote at these listed points.

Show worked solution →

Part (a): divide $f$ by $g$ at each listed $x$ .

\frac{f(-2)}{g(-2)} = \frac{3}{-1} = -3, \quad \frac{f(-1)}{g(-1)} = \frac{0}{1} = 0, \quad \frac{f(0)}{g(0)} = \frac{-1}{2} = -\tfrac{1}{2},

\frac{f(1)}{g(1)} = \frac{0}{1} = 0, \quad \frac{f(2)}{g(2)} = \frac{3}{-1} = -3.

So the quotient takes the values $-3,\; 0,\; -\tfrac{1}{2},\; 0,\; -3$ at $x = -2, -1, 0, 1, 2$ .

Part (b): intercepts from zeros of $f$ ; asymptotes from zeros of $g$ . The quotient is zero where $f = 0$ with $g \neq 0$ , namely at $x = -1$ and $x = 1$ , so those are the $x$ -intercepts. None of the listed points has $g = 0$ (the $g$ -values are $-1, 1, 2, 1, -1$ ), so there is no division by zero and hence no vertical asymptote at any of them.

Marker's note: two marks for the five correct quotient values, one for the intercepts at $x = \pm 1$ , one for explaining that no listed $g$ -value is zero. Marking $x = 0$ (where $f = -1$ , not $0$ ) as an intercept is the trap.

core3 marksThe function

f

is positive everywhere and has a local maximum at

(2, 4)

and a local minimum at

(5, 1)

. Describe what happens to each of these two turning points on the graph of

y = \dfrac{1}{f(x)}

, giving the coordinates of the resulting points.

Show worked solution →

Reciprocating flips the size ordering where $f > 0$ . A large positive value becomes a small positive value and vice versa, so a maximum of $f$ becomes a minimum of $\dfrac{1}{f}$ , and a minimum of $f$ becomes a maximum of $\dfrac{1}{f}$ .

Transform each turning point. The maximum of $f$ at $(2, 4)$ becomes a local minimum of $\dfrac{1}{f}$ at

\left(2,\ \tfrac{1}{4}\right).

The minimum of $f$ at $(5, 1)$ becomes a local maximum of $\dfrac{1}{f}$ at

\left(5,\ \tfrac{1}{1}\right) = (5,\ 1).

Marker's note: one mark for the max-becomes-min flip with $\left(2, \tfrac14\right)$ , one for the min-becomes-max flip, one for the reciprocated $y$ -values. Keeping the $y$ -values unchanged (not reciprocating them) is the common error.

exam5 marksA machine part vibrates so that its displacement from rest is modelled by

y = h(t) = e^{-t}\sin(2t)

centimetres, for

t \ge 0

seconds. (a) Show that the graph oscillates between the envelope curves

y = e^{-t}

and

y = -e^{-t}

, and state the exact times in

0 \le t \le \pi

at which

y = 0

. (b) Show that

h(t)

touches the upper envelope

y = e^{-t}

whenever

\sin(2t) = 1

, and find the first such time. (c) Explain, using the envelope, why the vibration dies away, and state the displacement at

t = 0

Show worked solution →

Part (a): bound the oscillating factor. For all $t$ , $-1 \le \sin(2t) \le 1$ . Multiplying through by $e^{-t}$ , which is positive, preserves the inequalities:

-e^{-t} \le e^{-t}\sin(2t) \le e^{-t},

so the graph is squeezed between $y = e^{-t}$ and $y = -e^{-t}$ , the envelope curves. The displacement is zero where the bounded factor is zero: $\sin(2t) = 0$ gives $2t = 0, \pi, 2\pi$ , so on $0 \le t \le \pi$ ,

t = 0, \quad t = \frac{\pi}{2}, \quad t = \pi.

Part (b): touching the upper envelope. The graph meets $y = e^{-t}$ exactly when $e^{-t}\sin(2t) = e^{-t}$ , and since $e^{-t} \neq 0$ this requires

\sin(2t) = 1.

The first solution for $t \ge 0$ is $2t = \dfrac{\pi}{2}$ , giving

t = \frac{\pi}{4} \text{ s}.

Part (c): why it decays, and the start value. The envelope $y = \pm e^{-t}$ shrinks towards zero as $t$ increases, because $e^{-t} \to 0$ . Since the displacement is trapped between the two envelope curves, its swings must shrink with them, so the vibration dies away. At the start,

h(0) = e^{0}\sin(0) = 1 \times 0 = 0 \text{ cm}.

Marker's note: one mark for bounding $\sin(2t)$ and deriving the envelope, one for the three zeros in (a); one for reducing the touch condition to $\sin(2t) = 1$ and one for $t = \tfrac{\pi}{4}$ in (b); one for the decay explanation with $h(0) = 0$ in (c). Solving $\sin(2t) = 1$ as $2t = \tfrac{\pi}{2}$ (not $t = \tfrac{\pi}{2}$ ) is the step most often botched.

exam5 marksLet

f(x) = x^2 - 4

. (a) Sketch the key features of

y = \dfrac{1}{f(x)}

, stating the equations of all asymptotes. (b) Show that

\dfrac{1}{f(x)}

has a turning point at

x = 0

and find its coordinates. (c) Hence state, with reasons, the range of values

k

for which the horizontal line

y = k

meets

y = \dfrac{1}{f(x)}

at no points in the region between the vertical asymptotes.

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Part (a): read the reciprocal off $f(x) = x^2 - 4$ . The parent has zeros at $x = \pm 2$ and a minimum at $(0, -4)$ . Zeros of $f$ become vertical asymptotes:

x = -2 \quad \text{and} \quad x = 2.

As $|x| \to \infty$ , $f(x) \to \infty$ , so $\dfrac{1}{f(x)} \to 0$ , giving the horizontal asymptote $y = 0$ . Outside the asymptotes $f > 0$ so $\dfrac{1}{f} > 0$ (small, near the axis); between them $f < 0$ so $\dfrac{1}{f} < 0$ , dropping to $-\infty$ near each asymptote.

Part (b): the turning point comes from the minimum of $f$ . The minimum of $f$ at $(0, -4)$ , where $f < 0$ , reciprocates to a local maximum of $\dfrac{1}{f}$ at

\left(0,\ \frac{1}{-4}\right) = \left(0,\ -\tfrac{1}{4}\right).

Confirm by calculus: with $y = (x^2 - 4)^{-1}$ , $\dfrac{dy}{dx} = -\dfrac{2x}{(x^2 - 4)^2}$ , which is zero at $x = 0$ , and the middle branch peaks there, so $\left(0, -\tfrac14\right)$ is a maximum.

Part (c): use the shape of the middle branch. Between the asymptotes the branch runs from $-\infty$ up to its highest point $y = -\tfrac{1}{4}$ at $x = 0$ and back down to $-\infty$ . So its output covers every value with $y \le -\tfrac{1}{4}$ , and never rises above $-\tfrac{1}{4}$ . A line $y = k$ misses this branch entirely exactly when it sits above the peak:

k > -\frac{1}{4}.

Marker's note: one mark for the vertical asymptotes $x = \pm 2$ and $y = 0$ , one for the correct sign of each branch in (a); one for the flipped turning point $\left(0, -\tfrac14\right)$ in (b); one for identifying $-\tfrac14$ as the greatest value on the middle branch and one for the range $k > -\tfrac14$ in (c). Forgetting that the middle branch is entirely negative (and so testing $k$ against $+\tfrac14$ ) is the trap.