Graph Transformations: past paper questions & worked solutions

For $y=f(2x)$ the horizontal coordinate is divided by 2 while the vertical coordinate is unchanged: $(-4,-3)\mapsto(-4/2,-3)$.

The inside shift moves the point right by 1; the outside factor multiplies its ordinate by 3. Thus $(-4,-3)\mapsto(-4+1,3(-3))$.

Output modulus retains the input and replaces the output by its absolute value: $(-4,-3)\mapsto(-4,|-3|)$.

The modulus is smallest when $x-5=0$. At $x=5$, $f(5)=10$.

For $x\lt 5$, $|x-5|=5-x$, so

This is consistent with $x\lt 5$. For $x\ge5$, $|x-5|=x-5$, so $2x\gt 6x$, which requires $x\lt 0$ and is inconsistent with this branch. The equality point $x=5/2$ is excluded because the inequality is strict.

The transformation $y=3f(x-2)$ moves a point $(x,y)$ to $(x+2,3y)$. Thus $P=(5,10)$ moves to $(5+2,3\cdot10)$.

Since $x\ge0$, $2x^2\ge0$. The minimum is $f(0)=-5$ and there is no upper bound.

Reflect the given curve in $y=x$. Solving $x=2y^2-5$ with $y\ge0$ gives $y=\sqrt{(x+5)/2}$, $x\ge-5$. The inverse starts at $(-5,0)$, passes through $(0,\sqrt{5/2})$, and is increasing and concave down. The solid inverse curve below is paired with the dashed original curve.

Here $f$ is strictly increasing on $[0,\infty)$, so an intersection of $f$ and its inverse must be on $y=x$: if $y=f(x)\gt x$, then $f(y)\gt f(x)=y$, contradicting $f(y)=x$; the case $y\lt x$ is similar. Hence

Reject the negative root because the original curve has $x\ge0$.

Expand $R\cos(x-\alpha)=R\cos\alpha\cos x+R\sin\alpha\sin x$. Comparing coefficients gives $R\cos\alpha=1$ and $R\sin\alpha=2$. Thus $R^2=1^2+2^2=5$ and $\tan\alpha=2$. The specified first-quadrant phase is $\alpha=\arctan2=1.1071487\ldots$ radians.

Using the exact phase $\alpha=\arctan2$, write $g(x)=3-7\sqrt5\cos(2x-\alpha)$. (i) The negative cosine coefficient means the maximum occurs when the cosine is $-1$; the maximum is $3+7\sqrt5$. (ii) $2x-\alpha=\pi+2n\pi$, so $x=(\pi+\alpha)/2+n\pi$. The $n=-1$ value is negative; $n=0$ is the smallest positive one, $x=2.1243706857\ldots$, or 2.12 to two decimal places.

(i) Extraction starts at $G=0$: $30e^{1-0.05k}=40$, so $1-0.05k=\ln(4/3)$ and $k=20(1-\ln(4/3))=14.24635855\ldots$ years. (ii) This is 14 years and $12(0.24635855\ldots)=2.95630261\ldots$ months after 1 January 1800. Two full months have elapsed from January 1814, so the date is in March 1814. Use the unrounded fractional year to identify the month.

On 1 January 1870, $t=70$. Substitute: $G(70)=40-30e^{1-0.05(70)}=40-30e^{-2.5}=37.53745004\ldots$ tonnes.

As $t\to\infty$, $e^{1-0.05t}\to0$, so $G\to40$ tonnes.

For $y=3f(2x)$, a source point $(u,v)$ maps to $(u/2,3v)$. Thus the y-intercept stays $(0,0)$, the maximum becomes $(1/2,6)$ and the minimum becomes $(3/2,0)$. The horizontal compression and positive vertical stretch preserve the turning-point types. The sketch below records the given shape and mapped features; it does not assume an additional formula for f.

For $y=f(-x)-1$, map $(u,v)$ to $(-u,v-1)$: reflect in the y-axis, then shift down 1. The y-intercept is $(0,-1)$, the maximum is $(-1,1)$ and the minimum is $(-3,-1)$. The left end now rises and the right end falls.

Set $u=|x|\ge0$. Then

Reject $u=-3$ since $u\ge0$. Therefore $|x|=8$.

For $x\le0$, the right side $\frac52x\le0$ and the left side is nonnegative, so every such $x$ satisfies the inequality. For $0\lt x\lt 5$, $f(x)\lt 0$ and

For $x\ge5$, $f(x)\ge0$ and

Combining the cases retains equality at both boundary points.

For $x\ge0$, $0\lt 4/(3x+2)\le2$. Thus $f(0)=3$ is attained, while $f(x)$ increases towards 5 without reaching it.

(i) Put $y=5-4/(3x+2)$. Rearranging gives $(5-y)(3x+2)=4$, hence

Exchange variables. (ii) The inverse domain is the range of $f$ found above.

The original function has an absolute value and angle $x/3+\pi/6$:

Then $fg(-\pi)=f(2)=5-4/(3\cdot2+2)=5-1/2=9/2$.

The principal arcsine curve increases from $(-2,-\pi/2)$ through $(0,0)$ to $(2,\pi/2)$. It is concave down on the negative-x half, concave up on the positive-x half, and vertical at both ends.

Differentiate the supplied inverse relation: $dx/dy=2\cos y$. For $-\pi/2\lt y\lt \pi/2$, cosine is positive, so $\cos y=\sqrt{1-\sin^2y}=\sqrt{1-x^2/4}$. Taking the reciprocal gives

Thus $A=4$; the endpoints have vertical tangents, not finite derivatives.

At $y=\pi/4$, $x=2\sin(\pi/4)=\sqrt2$. The gradient is $1/\sqrt{4-2}=1/\sqrt2$. Hence $y-\pi/4=(x-\sqrt2)/\sqrt2$, so $y=x/\sqrt2+\pi/4-1$.

Expanding the required form gives

Comparing coefficients with \(5\cos\theta-4\sin\theta\),

Squaring and adding, using \(R\gt 0\), gives

Also, since \(0\lt \alpha\lt \frac\pi2\),

Thus the exact form is

A stretch parallel to the \(y\)-axis with scale factor \(\boxed{\sqrt{41}}\), taking \(y=\cos\theta\) to \(y=\sqrt{41}\cos\theta\).

A translation by the vector

This moves the curve \(\alpha\) units to the left, giving \(y=\sqrt{41}\cos(\theta+\alpha)\).

Since \(\theta\) can be any real number,

Hence \(4\leq4+(f(\theta))^2\leq45\). Taking reciprocals of these positive bounds and multiplying by \(90\) gives

Both endpoints are attained: the minimum occurs when \(f(\theta)=\pm\sqrt{41}\), and the maximum when \(f(\theta)=0\).

Using the quotient rule,

Thus $a=4$, $b=-24$ and $c=27$.

A stationary point requires

The roots are $3/2$ and $9/2$. Reject $9/2$ because the stated domain is $x\lt 3$. At $x=3/2$, $f(x)=-2$, so the turning point is $\boxed{(3/2,-2)}$. The derivative changes from positive to negative there, confirming a maximum.

The inside factor 2 halves the horizontal coordinate; adding 3 raises the vertical coordinate by 3. Therefore

Let $y=(4x+3)/(x-2)$. Then $yx-2y=4x+3$, so

Swap the input and output names. Since $f(x)=4+11/(x-2)$ can never equal 4, the inverse domain excludes 4.

Substitute $f(x)$ into both occurrences of the argument:

Thus $a=19$, $b=6$, $c=2$, $d=7$. The composite retains $x\ne2$ from the inner function and also requires $f(x)\ne2$, that is $x\ne-7/2$.

For $y=2f(3x)+8$, the point mapping is $(x,y)\mapsto(x/3,2y+8)$. Hence $(3,15)\mapsto(3/3,2(15)+8)=(1,38)$.

For $x\le a/3$, $y=a-3x$; for $x\ge a/3$, $y=3x-a$. These two straight branches form a V with minimum $(a/3,0)$ and $y$-intercept $(0,a)$.

Subtracting $b$ translates the previous graph vertically down by $b$. The minimum becomes $(a/3,-b)$ and the $y$-intercept becomes $(0,a-b)$, which is above the $x$-axis because $a\gt b\gt 0$.

For $x\le a/3$, the equation becomes $a-3x-b=5x$, giving $x=(a-b)/8$. This is positive and less than $a/8\lt a/3$, so it satisfies this branch condition. For $x\ge a/3$, the equation becomes $3x-a-b=5x$, giving $x=-(a+b)/2\lt 0$. This contradicts $x\ge a/3\gt 0$ and must be rejected. Therefore only one solution remains.

Since $x\gt 0$, $6\sqrt{x}$ is positive. Therefore $f(x)=0$ requires $\ln(4x)=0$, so $4x=1$ and $x=1/4$.

Apply the product and chain rules:

At $Q$, set $f'(x)=0$. Since $3/\sqrt{x}\gt 0$,

Then $\sqrt{x}=1/(2e)$ and $\ln(4x)=-2$, giving $f(x)=6(1/(2e))(-2)=-6/e$.

The factor $3/\sqrt{x}$ is positive and $\ln(4x)+2$ increases through zero at $Q$, so $f'$ changes from negative to positive: $Q$ is the global minimum. As $x\to\infty$, $f(x)\to\infty$, hence the range of $f$ is $[-6/e,\infty)$. Multiplication by $-2$ reverses the inequality and includes the maximum $12/e$.

For $y=|3x-5a|-2a$ with $a\gt 0$: (i) At $x=0$, $y=5a-2a=3a$, so $P=(0,3a)$. (ii) At $y=0$, $|3x-5a|=2a$ gives $3x-5a=\pm2a$. Thus $x=a$ or $7a/3$; since $a\lt 7a/3$, $Q=(a,0)$ and $R=(7a/3,0)$. (iii) The modulus vanishes at $x=5a/3$, giving the minimum $S=(5a/3,-2a)$.

Split $|3x-5a|-2a=|x-2a|$ at $x=5a/3$ and $x=2a$; their order uses $a\gt 0$. For $x\lt 5a/3$: $3a-3x=2a-x$, so $x=a/2$, which is in this region. For $5a/3\le x\lt 2a$: $3x-7a=2a-x$, so $x=9a/4$; this is outside the assumed region and is rejected. For $x\ge2a$: $3x-7a=x-2a$, so $x=5a/2$, which is valid. These three regions cover the whole real line, so there are exactly two solutions.

(i) $y=k-2|x|$ is an inverted V with vertex $(0,k)$ and roots $x=\pm k/2$. (ii) $y=|2x-k/3|$ is an upright V with vertex $(k/6,0)$ and y-intercept $(0,k/3)$. Because $k\gt 0$, both sketches have the ordering shown. Axes below use coordinates normalized by k, while labels give the original coordinates.

\qquad

Consider the breakpoints $0$ and $k/6$. For $x\lt 0$, the equation is $k/3-2x=k+2x$, giving $x=-k/6$, valid in this branch. For $0\le x\lt k/6$, it is $k/3-2x=k-2x$, impossible because $k\gt 0$. For $x\ge k/6$, it is $2x-k/3=k-2x$, giving $x=k/3$, valid in this branch. Both have nonnegative right side and satisfy the original equation.

Expand $R\sin(x-\alpha)=R\cos\alpha\sin x-R\sin\alpha\cos x$. Therefore $R\cos\alpha=12$, $R\sin\alpha=5$. Thus $R=\sqrt{12^2+5^2}=13$ and $\alpha=\arctan(5/12)=0.3947911197\ldots$ radians, in the required first quadrant.

Write $g(\theta)=10+13\sin(2\theta-\pi/6-\alpha)$. (i) The minimum is $10-13=-3$. (ii) For sine to equal $-1$, $2\theta-\pi/6-\alpha=-\pi/2+2\pi n$. Hence $\theta=-\pi/6+\alpha/2+\pi n$. For $n=0$ this is negative, so the smallest permitted value uses $n=1$:

Use the unrounded phase angle for the calculation.

As $\beta$ varies over the reals, $12\sin\beta-5\cos\beta$ covers $[-13,13]$. Its square therefore covers $[0,169]$, not $[169,169]$. Subtracting from 10 reverses the bounds, giving $[-159,10]$; both endpoints are attained.

Apply f twice, in order: $f(6)=21-2|2-6|=13$. Then $ff(6)=f(13)=21-2|2-13|=21-22=-1$. The intermediate input 13 lies in the domain $x\ge0$.

For $0\le x\le2$, $f(x)=17+2x$, so $17+2x=5x$ gives $x=17/3$, outside that branch; reject it. For $x\ge2$, $f(x)=25-2x$. Then $25-2x=5x$, giving $x=25/7$. This belongs to the second branch and satisfies the original modulus equation.

On the left branch $0\le x\le2$, f rises from $f(0)=17$ to $f(2)=21$. The right branch decreases from 21 without lower bound. A horizontal line therefore meets both branches at distinct points exactly when $17\le k\lt 21$. At $k=17$, one root is the permitted endpoint $x=0$; at $k=21$, the two intersections merge at the vertex, so 21 is excluded.

The original vertex is $(2,21)$. Under $y=af(x-b)$ it maps to $(2+b,21a)$. Comparing with $(6,3)$ gives $2+b=6$ and $21a=3$.

Use $2\sin x\cos x=\sin2x$ and $2\sin^2x=1-\cos2x$:

Expanding $R\sin(2x+\alpha)$ and comparing coefficients gives

Therefore

Thus $h(x)=5+\frac{3\sqrt5}{2}\sin(2x+\alpha)$ with the exact angle $\alpha=\arctan(2)$.

The given interval includes two complete periods. The range is

At a minimum, $\sin(2x+\alpha)=-1$, so

On $0\le x\le2\pi$, $n=0$ and $n=1$ give the two minima. The larger coordinate is

At $t=0$, $e^{-0.05t}=1$, so $H=280+24=304$ degrees Celsius.

The curve starts at $(0,304)$ and decreases for $t\ge0$. The exponential term stays positive and tends to zero, so the curve approaches $H=24$ from above without crossing it.

Substitute $H=144$:

Take natural logarithms: $-0.05t=\ln(3/7)$. Hence $t=-20\ln(3/7)=20\ln(7/3)=16.9459572077\ldots$ minutes.

Differentiate using the chain rule:

The original equation gives $280e^{-0.05t}=H-24$, so

Use $\sin2x=2\sin x\cos x$ and $2\cos^2x=1+\cos2x$:

Expanding $R\sin(2x+\alpha)$ and matching coefficients gives

Squaring and adding, $R^2=20$, so $R=2\sqrt5$ since $R\gt 0$. Also $\tan\alpha=1/2$, and $0\lt \alpha\lt \pi/2$ selects $\alpha=\arctan(1/2)=0.463647609\ldots$ radians. Thus $f(x)=2\sqrt5\sin(2x+\arctan(1/2))-1$. With the phase displayed to 3 significant figures, $f(x)\approx2\sqrt5\sin(2x+0.464)-1$.

(i) The sine term has maximum 1, hence the maximum of $f$ is $2\sqrt5-1$. (ii) Maxima occur when

The value for $k=0$ is $0.553574\ldots\gt 0$, while $k=-1$ is negative. The second positive value therefore has $k=1$:

The logarithm requires $x+1\gt 0$, so the domain is $x\gt -1$. As $x\to-1^+$, $\ln(x+1)\to-\infty$ and the ordinate tends to infinity.

(i) At the $y$-axis, $x=0$, giving $y=|2-4\ln1|=2$.

(ii) At the $x$-axis, $|2-4\ln(x+1)|=0$, so

Thus $A=(0,2)$ and $B=(e^{1/2}-1,0)$.

On $x\gt -1$, split the absolute-value inequality into two alternatives:

Dividing by $-4$ reverses the inequality in each branch:

Since the exponential is increasing, this becomes $x+1\lt e^{-1/4}$ or $x+1\gt e^{5/4}$. Intersect with the logarithmic domain; neither boundary is included because the source inequality is strict.

The modulus is smallest when $x-2=0$, giving $x=2$ and $f(2)=-10$. The two straight-line branches meet there, so this is the vertex of the V-shaped graph.

First $f(0)=3|0-2|-10=-4$. Then $ff(0)=f(-4)=3|-4-2|-10=18-10=8$. In particular $ff(0)$ means $f(f(0))$, not the square of $f(0)$.

For $x\lt 2$, the inequality becomes $3(2-x)-10\lt 5x+10$, so $-4-3x\lt 5x+10$, giving $x\gt -7/4$. This yields $-7/4\lt x\lt 2$. For $x\ge2$, it becomes $3x-16\lt 5x+10$, giving $x\gt -13$; every $x\ge2$ satisfies this. The union is $x\gt -7/4$.

Let $u=|x|\ge0$. Then $f(|x|)=0$ gives

Thus $u=16/3$ or $u=-4/3$. Reject $u=-4/3$ because $u\ge0$; $|x|=16/3$ has the two solutions below.

The product rule gives

Comparison with the stated form gives $A=2$.

Since $e^{-4x}\gt 0$, $f'(x)=0$ means $2x+1=0$ or $1-4x=0$. Thus $x=-1/2$ or $x=1/4$. Substitution gives $f(-1/2)=0$ and $f(1/4)=(3/2)^3e^{-1}=27/(8e)$.

The derivative changes from positive to negative at $x=1/4$, so this is the maximum; at $x=-1/2$ it does not change sign. The transformation $g(x)=8f(x-2)$ sends $(x,y)$ to $(x+2,8y)$. Hence the maximum stationary point is $(1/4+2,8\cdot27/(8e))=(9/4,27/e)$.

Expand $R\sin(x-\alpha)=R\cos\alpha\sin x-R\sin\alpha\cos x$. Thus $R\cos\alpha=8$, $R\sin\alpha=15$, so $R=\sqrt{64+225}=17$ and $\alpha=\arctan(15/8)=1.0808390005\ldots$ radians, in the required quadrant.

Using the exact phase from (a),

The denominator lies between 7 and 75, so it is always positive. (i) The smallest fraction occurs at its largest denominator: $f_{\min}=15/75=1/5$. (ii) This occurs when $x-\alpha=\pi/2+2n\pi$. The smallest positive value is $x=\alpha+\pi/2=2.6516353273\ldots$, or 2.65 to three significant figures.

The transformation $2f(x)-5$ preserves the locations of minima and maps their y coordinate to $2(1/5)-5=-23/5$.

Reflection in the x-axis changes minima of f into maxima of $-f$. The argument $2x$ halves their x coordinates, so the first positive maximum is at

At the vertex the inner expression vanishes: $3x+a=0$, giving $x=-a/3$ and $f(x)=a$.

$g(x)=|x+5a|$ is a V translated left by $5a$, with vertex $(-5a,0)$ and y-intercept $(0,5a)$. Since $a\gt 0$, the vertex lies strictly left of the y-axis. Coordinates in the drawing are normalized by a; labels give the actual intercepts.

Solve $|x+5a|=|3x+a|+a$ using breakpoints $-5a\lt -a/3$. For $x\lt -5a$: $-x-5a=-3x$, giving $x=5a/2$, outside this branch, so reject it. For $-5a\le x\lt -a/3$: $x+5a=-3x$, giving $x=-5a/4$ and $y=15a/4$. For $x\ge-a/3$: $x+5a=3x+2a$, giving $x=3a/2$ and $y=13a/2$. The last two values belong to their respective intervals and satisfy both original functions.

At P, $18=5e^{x-1}+3$, so $e^{x-1}=3$. Taking natural logarithms gives $x-1=\ln3$, hence $x=1+\ln3=\ln e+\ln3=\ln(3e)$.

Define $F(x)=5e^{x-1}+x^2-7$; its zeros are the intersection abscissae, and F is continuous. At the endpoints of the three-decimal rounding interval,

Continuity guarantees a root between them. Moreover $F'(x)=5e^{x-1}+2x\gt 0$ on this positive interval, so that root is unique there. It is the positive intersection $\alpha$, and rounds to 1.134 to three decimal places.

Start at $x_1=-3$ and keep full precision in $x_{n+1}=-\sqrt{7-5e^{x_n-1}}$:

The later iterates agree to six decimal places. The negative root corresponds to $\beta$ in the figure.

Expand $R\cos(x-\alpha)=R\cos\alpha\cos x+R\sin\alpha\sin x$. Matching coefficients gives $R\cos\alpha=1$, $R\sin\alpha=4$. Thus $R=\sqrt{1^2+4^2}=\sqrt{17}$ and $\tan\alpha=4$. Since $0\lt \alpha\lt \pi/2$, $\alpha=\arctan4=1.32581766\ldots$ radians.

The model becomes $H=24/[3+\sqrt{17}\cos(t/2-\alpha)]$. For $0\le t\le6.5$, the phase lies between $-\alpha$ and $3.25-\alpha$; throughout this interval the denominator remains positive. Its maximum $3+\sqrt{17}$ is attained at $t=2\alpha=2.651635\ldots$, inside the time interval. The height is then smallest:

To the nearest centimetre this is $3.37\ \mathrm m$ (337 cm).

For $H=10$, $3+\sqrt{17}\cos(t/2-\alpha)=24/10$, so

Let $\gamma=\arccos[-3/(5\sqrt{17})]$. In the permitted phase interval $[-\alpha,3.25-\alpha]$, only $+\gamma$ occurs: $-\gamma\lt -\alpha$, and shifts by $2\pi$ are outside it. Using the unrounded $\alpha=\arctan4$,

At the vertex, the expression inside the modulus is zero:

Hence \(\boxed{P=\left(\frac52,3\right)}\).

Write the function in its two branches:

On the left branch,

This lies in \(0\leq x\leq\frac52\), so it is valid. On the right branch,

This satisfies \(x\geq\frac52\), so it is also valid. Thus

Each straight branch can have at most one intersection with \(y=kx+2\). Two distinct roots therefore require one intersection strictly to each side of the vertex.

For the left branch,

The condition \(0\leq x_L\lt \frac52\) first requires \(k+4\gt 0\), then gives

For the right branch,

The condition \(x_R\gt \frac52\) requires \(k\lt 4\), and then

At \(k=\frac25\), both intersections coincide at the vertex, giving only one root. At \(k=4\), the line is parallel to the right branch and has no intersection with that branch, again leaving only one root. Hence

At the vertex, $3x-11=0$, so $P=(\frac{11}{3},-4)$.

Thus $3x-11=12$ or $3x-11=-12$, giving $x=\frac{23}{3}$ or $x=-\frac13$.

Write the two branches explicitly:

On the left branch, $mx=7-3x$ gives $x=7/(m+3)$ for $m\ne-3$. This lies on the left branch precisely when $m\lt -3$ or $m\ge-12/11$. On the right branch, $mx=3x-15$ gives $x=15/(3-m)$ for $m\ne3$. This lies on the right branch precisely when $-12/11\le m\lt 3$. At $m=-12/11$, the two branch intersections coincide at the vertex, so there is only one distinct point. For $m\lt -3$ or $m\ge3$, only the left intersection exists. At $m=-3$ there is no intersection; at $m=3$ there is one. Consequently the complete answer is

Under $y=af(x-b)$, the vertex becomes $(\frac{11}{3}+b,-4a)$. Equating this with $(5,16)$ gives

For $x\gt -7/4$, apply the product and chain rules:

The interior stationary values satisfy $x^2(2x+3)=0$, so $x=0$ or $x=-3/2$. Near $x=-3/2$, the derivative changes from negative to positive; near $x=0$ it stays positive on both sides. Thus the minimum turning point is at $x=-3/2$.

The endpoint has $f(-7/4)=0$, the global minimum is $-27/8$ and $f(x)$ tends to infinity as $x$ increases. Hence the range of $f$ is $[-27/8,\infty)$. Multiplication by $-4$ reverses the inequality and makes the upper endpoint $27/2$.

The transformation is a shift right by $3/2$, vertical stretch by 40, then shift down by 8:

Therefore $Q=(1/2,3/8)$ maps to $(1/2+3/2,\;40\cdot3/8-8)$.

Use product and chain rules:

With a common denominator,

This derivative formula applies for $x\gt -9/4$; the endpoint derivative is not finite.

The denominator is positive in the derivative domain. Thus $5x^2+9x-2=(5x-1)(x+2)=0$.

The derivative changes from positive to negative at $x=-2$, so this is the local maximum P. Substitute into f: $f(-2)=5(4-2)\sqrt{-8+9}=10$.

On $[-9/4,0]$, f rises from $f(-9/4)=0$ to $f(-2)=10$, then decreases to $f(0)=-30$. The other stationary point $1/5$ is outside this restricted interval. Thus the minimum and maximum of f on it are $-30$ and 10. Applying the positive vertical transformation $g=2f+4$ gives endpoints $2(-30)+4=-56$ and $2(10)+4=24$, both attained.

At the vertex $2x+7=0$, hence $x=-7/2$ and $f(x)=-10$.

For $x\ge-7/2$, $f(x)=x-13/2$. The inequality becomes $x-13/2\ge x/3+1$, giving $x\ge45/4$, which satisfies this branch. For $x\lt -7/2$, $f(x)=-x-27/2$. Then $-x-27/2\ge x/3+1$, giving $x\le-87/8$, which satisfies that branch. Join the two permitted outer intervals, including the equality endpoints.

The zeros of f satisfy $|2x+7|=20$, giving $x=-27/2$ and $x=13/2$. Reflect the part below the x-axis upward, leaving the positive branches unchanged. This creates minima at $(-27/2,0)$ and $(13/2,0)$ and a local maximum at $(-7/2,10)$.

The modulus vanishes at $3x-13=0$, giving $x=13/3$ and $y=5$.

(i) The modulus is nonnegative and attains zero, so the range is $[5,\infty)$. (ii) $f(4)=|12-13|+5=6$, then $ff(4)=f(6)=|18-13|+5=10$.

For $x\lt 13/3$, $|3x-13|+5=18-3x$. Thus $16-2x\gt 18-3x$ gives $x\gt 2$, so this branch contributes $2\lt x\lt 13/3$. For $x\ge13/3$, the right side is $3x-8$. Then $16-2x\gt 3x-8$ gives $x\lt 24/5$, so this branch contributes $13/3\le x\lt 24/5$. Combine the adjacent intervals.

The new vertex is at $(13/3-b,5a)$. Therefore $5a=20$ and $13/3-b=4$, giving $a=4$ and $b=1/3$.

Substitute $g(x)=5-2x^2$ into f:

The composite requires $x\le0$ from g and $g(x)\ne5$ from f; therefore $x\lt 0$. Now $4-13/x^2=3$ gives $x^2=13$. Only the negative root belongs to the composite domain.

Let $y=(4x+6)/(x-5)$. Then $yx-5y=4x+6$, so $x(y-4)=5y+6$ and $x=(5y+6)/(y-4)$. Exchange the variable names. The original function is $4+26/(x-5)$, whose range excludes 4.

The graph of g is only the left half of the downward parabola: $y=5-2x^2$, $x\le0$. Its intercepts are $(0,5)$ and $(-\sqrt{5/2},0)$. For the inverse, $x=5-2y^2$ and $y\le0$, hence $g^{-1}(x)=-\sqrt{(5-x)/2}$, $x\le5$. Reflect the original branch in $y=x$. The inverse intercepts are $(5,0)$ and $(0,-\sqrt{5/2})$.

(i) Since $|2x-b|\ge0$, the maximum occurs at $2x-b=0$: $x=b/2$ and $y=a$. (ii) At $x=0$, $y=a-|-b|=a-b$ because $b\gt 0$. (iii) At an x-axis crossing, $|2x-b|=a$, hence $2x-b=\pm a$. Since $a\gt b\gt 0$, $(b-a)/2\lt 0$ and $(b+a)/2\gt 0$, matching the sketch.

The graph $y=|x|-1$ is a V with vertex $(0,-1)$ and intercepts $(-1,0)$ and $(1,0)$. Its left branch is $y=-x-1$ for $x\lt 0$ and its right branch is $y=x-1$ for $x\ge0$. Add this V to the supplied axes; the original inverted V remains unchanged.

The grey original curve is schematic; the three labelled points belong to the added graph.

At $x=-3$, the common ordinate is $|-3|-1=2$. Since $b\gt 0$, $|-6-b|=6+b$, so $a-(6+b)=2$, giving $a=b+8$. At $x=5$, the common ordinate is $|5|-1=4$, so $a-|10-b|=4$. Substitute $a=b+8$ to obtain $|10-b|=b+4$. If $b\le10$, then $10-b=b+4$, giving $b=3$, $a=11$. If $b\gt 10$, then $b-10=b+4$, impossible. Both accepted values satisfy $a\gt b\gt 0$.

At $t=0$, $18=A-180e^0=A-180$. Hence $A=198$.

At $t=5$, $90=198-180e^{-5k}$, so $180e^{-5k}=108$ and $e^{-5k}=3/5$. Taking logarithms gives $-5k=\ln(3/5)$, hence $k=(1/5)\ln(5/3)$. Thus one suitable rational pair is $p=1/5$, $q=5/3$.

Use the unrounded k:

Round the temperature to three significant figures.

Differentiate with respect to time: $d\theta/dt=180ke^{-kt}$. At $t=9$ this is $180[\ln(5/3)/5]e^{-(9/5)\ln(5/3)}=7.332421555\ldots$. The sign is positive because the oven is warming.

(i) The modulus is zero at $3x-22=0$, giving the vertex $(22/3,5)$. (ii) At $x=0$, $y=5-22=-17$, so the y-axis intersection is $(0,-17)$.

Set $y=0$: $|3x-22|=5$. Thus $3x-22=5$ or $-5$, giving $x=9$ or $17/3$.

The curve $C_2:y=x^2/9-9$ is an upward-opening parabola, symmetric about the y-axis. Its vertex and y-intercept are $(0,-9)$; setting y=0 gives $x^2=81$, hence x-intercepts $(-9,0)$ and $(9,0)$. The sketch below includes $C_1$ for context and labels all three axis intersections of $C_2$.

For $x\le22/3$, $C_1$ is $y=3x-17$. Equating to $C_2$ gives

Only $x=3$ belongs to this branch; $y=-8$. For $x\ge22/3$, $C_1$ is $y=27-3x$. Then

Only $x=9$ belongs to this branch; $y=0$. The rejected candidates 24 and $-36$ lie outside their assumed branch regions.