3 Sequences, functions and graphs

Identify the starting position, apply the stated rule exactly, and label the terms. For a position rule, substitute $n=1,2,3,\ldots$ rather than repeatedly changing the previous value.

Check that every generated term obeys the same definition. A list may look familiar but still follow a different rule after its displayed terms.

For $5,9,13,17,\ldots$, each difference is $4$, so the next terms are $21,25$ and the term-to-term rule is ‘add 4’.

For $1,2,4,8,\ldots$, each term is twice the previous term, so the next terms are $16,32$ and the rule is ‘multiply by 2’.

For $7,11,15,19,\ldots$, $d=4$. The sequence $4n$ begins $4,8,12,16$, so add 3: $u_n=4n+3$.

To test whether 83 is a term, solve $4n+3=83$, giving $n=20$. Because $20$ is a positive integer, 83 is a term.

If $u_2=7$ and $u_5=19$, subtract $a+d=7$ from $a+4d=19$: $3d=12$, so $d=4$ and then $a=3$.

Rebuild the stated terms: $u_2=3+4=7$ and $u_5=3+16=19$. This catches an off-by-one error in the multiplier of $d$.

For $a=4$ and $d=3$, $u_{50}=4+49(3)=151$. The multiplier is 49 because there are 49 steps from the first term to the 50th.

If $4+(n-1)3=100$, then $3n+1=100$ and $n=33$, so 100 is the 33rd term. A non-positive or non-integer result would mean it is not a term.

For $4+7+10+13+\cdots$ with 50 terms, $a=4$, $d=3$, so $S_{50}=\frac{50}{2}[2(4)+49(3)]=25(155)=3875$.

The last term is $l=a+(n-1)d$. Writing the series forwards and backwards makes each pair total $a+l$; there are $n$ such pairs across two copies of the series.

A mapping diagram represents a function if every input has exactly one outgoing arrow. An output may receive several arrows, and some listed outputs may receive none.

The rule $x\mapsto x^2$ maps both $-3$ and $3$ to $9$; this is still a function because each input has only one output.

If $f(x)=3x-2$, then $f(5)=3(5)-2=13$ and $f(a+1)=3(a+1)-2=3a+1$.

Functions named $f$ and $g$ can use different rules. Keep the function name attached to the correct definition throughout a multi-part problem.

For $f(x)=1/(x-2)$, $x=2$ makes the denominator zero, so the domain excludes 2. Also $f(x)$ can never equal 0, so the range excludes 0.

On a graph, project the curve onto the $x$-axis for the domain and onto the $y$-axis for the range, respecting open endpoints, holes and asymptotes.

If $f(x)=2x+1$ and $g(x)=x^2$, then $fg(x)=2x^2+1$, while $gf(x)=(2x+1)^2$. Order matters.

For $f(x)=3x-5$, set $y=3x-5$ and solve $x=(y+5)/3$, so $f^{-1}(x)=(x+5)/3$.

A many-to-one rule such as $x^2$ needs a restricted domain, for example $x\ge0$, before its inverse $\sqrt{x}$ is a function.

To answer a question, locate the given value on one axis, move to the graph, then read the corresponding value from the other axis. For an interval, compare its two endpoints.

Check the scale on each axis separately: one square need not represent one unit, and the two axes may use different scales.

A point on the $x$-axis has $y=0$; a point on the $y$-axis has $x=0$. Such points are not in any quadrant.

After plotting, read the point back from the axes. This catches swapped coordinates and incorrect signs.

Mark known coordinates, use shape properties such as equal sides, parallel lines or perpendicular diagonals, and solve only for coordinates not already fixed.

For $A(-5,4)$ and $B(3,-2)$, $M=((-5+3)/2,(4-2)/2)=(-1,1)$.

The midpoint lies halfway in both directions, so average the two $x$-coordinates and independently average the two $y$-coordinates.

If 10 dollars converts to 60 krone, the rate is 6 krone per dollar. Thus 40 dollars corresponds to 240 krone.

Choose two clear, widely separated points on the line. Form a right-angled gradient triangle and divide the vertical change by the horizontal change.

For $2y=-7x+10$, divide every term by 2: $y=-3.5x+5$. The gradient is $-3.5$ and the intercept is $(0,5)$.

For a quadratic, use enough points around the turning point and look for symmetry. For a linear graph, two correct points determine the line, but a third point checks arithmetic.

For polynomial graphs, calculate a table including intercept regions and turning behaviour. For trigonometric graphs in degrees, mark key angles and repeat using the correct period.

Transform several defining points—intercepts, vertices and turning points—then preserve the original curve's connections and shape.

Use invariant features: vertical translations keep $x$-coordinates of turning points; horizontal transformations keep their $y$-coordinates.

Use points on the tangent, not necessarily points on the curve. A large gradient triangle reduces the effect of reading error.

To solve $x^2-5x=x-7$, plot $y=x^2-5x$ and $y=x-7$, or rearrange consistently to use an already drawn curve and a suitable line.

Between $(9,-4)$ and $(5,8)$, $m=(8-(-4))/(5-9)=12/(-4)=-3$.

Subtract coordinates in the same order on top and bottom. Reversing both orders gives the same gradient.

A line perpendicular to $y=-4x+5$ has gradient $1/4$. Through $(3,7)$ it satisfies $7=(1/4)(3)+c$, so $c=25/4$.

For a very small change in $x$, the corresponding change in $y$ is approximately $(dy/dx)\Delta x$. Differentiation gives this local gradient exactly for an algebraic function.

If $y=4x^3+5x^2+2x-7$, then $dy/dx=12x^2+10x+2$.

For $y=x^3-x^2-8x+12$, $dy/dx=3x^2-2x-8=(3x+4)(x-2)$. Thus stationary points occur at $x=-4/3$ and $x=2$.

Sketch or inspect the general shape around each stationary point. The curve's direction on both sides decides whether the point is a maximum or minimum.

‘Maximum’ or ‘minimum’ may mean local unless a domain is given. Endpoints must also be checked when finding an absolute extreme on a closed interval.

If $s=2t^3-5t^2+6t-5$, then $v=6t^2-10t+6$ and $a=12t-10$. Setting $a=5$ gives $t=1.25$ s.

Reject solutions outside the stated physical domain, such as negative time, impossible length or a value making a denominator zero.