Combining functions: sums, differences, products, quotients, squares and reciprocals

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.

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.

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:

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.

Part (a): add the two functions.

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:

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.

Square the linear factor.

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.

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

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

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 $x = \pi$:

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.

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

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.

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

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

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.

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:

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 $0 \le t \le \pi$,

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

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

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,

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.

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:

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

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:

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.