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Functions HL

IB Mathematics: Applications & Interpretation · Topic 2: Functions (HL)
www.jmaths.xyz
1
Composite Functions in Context
A composite function applies one function to the output of another — a chained model. The notation \( (f \circ g)(x) = f(g(x)) \) means “do \( g \) first, then \( f \)”.
Composition
\( (f \circ g)(x) = f\big(g(x)\big) \)
Inner function \( g \) acts first. In general \( f \circ g \neq g \circ f \).
Domain consideration
\( x \) must be valid for \( g \), and \( g(x) \) valid for \( f \)
e.g. if \( f(x)=\sqrt{x} \), then \( g(x) \geq 0 \) is required.
Real-world chaining
If \( c = g(n) \) gives cost from units, and \( p = f(c) \) gives profit from cost, then profit from units is \( (f \circ g)(n) = f(g(n)) \).
Units flow through the chain: feed the output of the first stage into the second.
TI-84 Plus CE
Store the inner function in Y1, the outer in Y2
On the home screen type Y2(Y1(3)) for \( (f\circ g)(3) \)
([VARS] → Y-VARS to paste \( Y1, Y2 \))
Or just evaluate the inner value first, then the outer.
TI-Nspire CX II
Define g(x):=... and f(x):=... with [ctrl][:=] or Define
Then type f(g(3)) on a Calculator page
For an algebraic form, type f(g(x))
Define functions once, reuse for any input.
Casio fx-CG50
[MENU] → Graph: enter Y1, Y2
On Run-Matrix type Y2(Y1(3)) (paste with [VARS] → GRAPH)
Or evaluate the inner value, then substitute
Keep the inner function in the lower Y-slot.
Worked Example
Let \( f(x)=2x+1 \) and \( g(x)=x^2 \). Find \( (f \circ g)(3) \).
Inner first: \( g(3)=3^2=9 \)
Then outer: \( f(9)=2(9)+1=19 \)
Answer: \( (f \circ g)(3)=19 \)
GDC: with \( Y1=x^2 \), \( Y2=2x+1 \), type Y2(Y1(3)) → 19.
Common error: doing \( g(f(x)) \) by mistake. The inner function (closest to \( x \)) is applied first: \( (f\circ g)(x)=f(g(x)) \), so \( g \) goes first, not \( f \).

Functions HL

IB Mathematics: Applications & Interpretation · Topic 2: Functions (HL)
www.jmaths.xyz
2
Transformations of Graphs
A transformation changes a graph’s position or shape. Changes outside the function (\( y \)) act vertically and behave as expected; changes inside the function (\( x \)) act horizontally and are reversed.
Translations
\( y = f(x) + b \): up by \( b \)
\( y = f(x - a) \): right by \( a \)
Vector \( \begin{pmatrix} a \\ b \end{pmatrix} \). Note \( f(x-a) \) shifts in the \( +x \) direction.
Reflections
\( y = -f(x) \): in the \( x \)-axis
\( y = f(-x) \): in the \( y \)-axis
Negative outside flips up/down; negative inside flips left/right.
Vertical stretch
\( y = p\,f(x) \): scale factor \( p \) (vertical)
\( y \)-values multiplied by \( p \); \( x \)-intercepts unchanged.
Horizontal stretch
\( y = f(qx) \): scale factor \( \dfrac{1}{q} \) (horizontal)
\( x \)-values divided by \( q \); \( y \)-intercept unchanged.
Composite transformations — effect on a point
Under \( y = p\,f(x - a) + b \), the point \( (x, y) \) maps to \( \big(x + a,\; p\,y + b\big) \).
Apply the horizontal stretch/shift to \( x \), the vertical stretch/shift to \( y \).
TI-84 Plus CE
Enter the original in Y1
Enter the transform referencing it, e.g. Y2 = 3*Y1(X-1)-2
[GRAPH] to compare both curves
Use [2ND][TRACE] → value to check mapped points.
TI-Nspire CX II
On a Graphs page define f1(x)=...
Then f2(x)=3·f1(x-1)-2 to plot the transform
Drag or use [menu] → Trace to read points
Reference \( f1 \) so both update together.
Casio fx-CG50
[MENU] → Graph: enter Y1
Enter Y2 = 3 Y1(X-1) - 2 (paste \( Y1 \) via [VARS] → GRAPH)
[F6] DRAW to overlay both
Use [SHIFT][F1] Trace to inspect.
Worked Example
The point \( (2, 5) \) lies on \( y = f(x) \). Find its image on \( y = 3f(x - 1) - 2 \).
Horizontal: shift right 1 ⇒ \( x: 2 \to 2 + 1 = 3 \)
Vertical: stretch \( \times 3 \) then down 2 ⇒ \( y: 5 \to 3(5) - 2 = 13 \)
Answer: image point \( (3, 13) \)
GDC: plot \( Y1 \) and \( Y2=3Y1(X-1)-2 \), trace to confirm.
Common error: reading horizontal shifts backwards. \( y = f(x - 1) \) moves the graph right by 1 (not left), and \( y = f(2x) \) compresses horizontally by factor \( \tfrac{1}{2} \) (not stretches). Inside the bracket, everything is reversed.

Functions HL

IB Mathematics: Applications & Interpretation · Topic 2: Functions (HL)
www.jmaths.xyz
3
Exponential Models & Half-Life
Exponential model with limit
\( f(x) = k\,a^{x} + c \)   or   \( f(x) = k\,e^{rx} + c \)
\( c \) is the horizontal asymptote (long-run limit). Growth if \( a > 1 \) (\( r > 0 \)); decay if \( 0 < a < 1 \) (\( r < 0 \)). Doubling time / half-life is found by solving for the time to double / halve.
Half-life & doubling (GDC-first): set the model equal to half (or double) the starting amount and solve by graph intersection — plot the model and the target line, then find the intersection \( x \).
TI-84 Plus CE
Fit data: [STAT] → CALC → ExpReg (gives \( y=a\cdot b^{x} \))
Solve: store model in Y1, target in Y2
[2ND][TRACE]5:intersect for the time
Or [MATH] Solver for \( Y1 = \text{value} \).
TI-Nspire CX II
Fit: [menu] → Statistics → Stat Calculations → Exponential Regression
Solve: nSolve(k·a^x + c = value, x)
Or graph the model and target, use Intersection Point(s)
nSolve returns the time directly.
Casio fx-CG50
Fit: [MENU] → Statistics → [F2] CALC → Exp regression
Solve: Graph the model and target line
[SHIFT][F5] G-Solv → INTSECT
Or use Equation → Solver for \( f(x)=\text{value} \).
Worked Example
A sample has mass \( M = 200(0.5)^{t/8} \) grams, \( t \) in hours. Find the half-life.
Half-life: solve \( 200(0.5)^{t/8} = 100 \Rightarrow (0.5)^{t/8} = 0.5 \)
So \( \dfrac{t}{8} = 1 \Rightarrow t = 8 \). (Check \( M=50 \): \( (0.5)^{t/8}=0.25 \Rightarrow t=16 \).)
Answer: half-life \( = 8 \) hours
GDC: intersect \( Y1=200(0.5)^{X/8} \) with \( Y2=100 \) → \( X=8 \).
Common error: ignoring the asymptote \( c \). For \( f(x)=k\,a^{x}+c \) the amount does NOT halve to a true half — halve the part above \( c \), i.e. solve \( f(x)=c+\tfrac{1}{2}(f(0)-c) \) when a vertical offset is present.
4
Logistic Models
Logistic model (given in formula booklet)
\( f(x) = \dfrac{L}{1 + C e^{-kx}} \),   \( L, C, k > 0 \)
\( L \) = carrying capacity (the upper limit as \( x \to \infty \)). Starting value \( f(0)=\dfrac{L}{1+C} \). \( k \) controls growth rate. Models population, spread of a virus, EV uptake.
TI-84 Plus CE
Fit data: [STAT] → CALC → Logistic
(returns \( y = c/(1 + a e^{-bx}) \), so \( L=c \))
Evaluate / solve via Y1 and intersect
Read \( L \) (carrying capacity) off the fit.
TI-Nspire CX II
Fit: [menu] → Statistics → Stat Calculations → Logistic Regression (d=0)
Read \( L \) as the numerator coefficient
Solve a time with nSolve(...) or graph intersection
Plot to see the S-curve flatten at \( L \).
Casio fx-CG50
Fit: [MENU] → Statistics → [F2] CALC → Logistic
The fit gives \( L \), \( C \), \( k \) (form \( L/(1+Ce^{-kx}) \))
Solve via G-Solv INTSECT
The horizontal asymptote is \( y = L \).
Worked Example
A population is modelled by \( P = \dfrac{10000}{1 + 49 e^{-0.5t}} \), \( t \) in years. State the carrying capacity and find \( P \) when \( t = 10 \).
Carrying capacity \( = L = 10000 \) (limit as \( t \to \infty \)).
\( P(10) = \dfrac{10000}{1 + 49 e^{-5}} = \dfrac{10000}{1 + 49(0.006738)} \approx \dfrac{10000}{1.3302} \approx 7518 \)
Answer: carrying capacity \( 10000 \); \( P(10) \approx 7518 \)
GDC: store the model in \( Y1 \), evaluate at \( X=10 \) → 7518.
5
Sinusoidal Models (Phase Shift)
Sinusoidal model
\( f(x) = a\sin\big(b(x - c)\big) + d \)
\( a \) = amplitude \( = \dfrac{\max - \min}{2} \);   period \( = \dfrac{2\pi}{b} \) (radians) or \( \dfrac{360^\circ}{b} \) (degrees);   \( c \) = horizontal (phase) shift;   \( d \) = principal axis \( = \dfrac{\max + \min}{2} \). Models tides, daylight, temperature, rotating blades.
TI-84 Plus CE
Fit data: [STAT] → CALC → SinReg (set RADIAN mode first)
Returns \( y = a\sin(bx + c) + d \)
Solve a time via Y1 & intersect
Amplitude \( =|a| \), period \( =2\pi/b \).
TI-Nspire CX II
Fit: [menu] → Statistics → Stat Calculations → Sinusoidal Regression
Set Angle = Radian in Document Settings first
Solve with nSolve(...) or graph intersection
Read \( a, b, c, d \) from the fit output.
Casio fx-CG50
Fit: [MENU] → Statistics → [F2] CALC → Sin regression
Set Angle = Rad in SET UP ([SHIFT][MENU]) first
Solve via G-Solv INTSECT
Period \( =2\pi/b \); axis \( y=d \).
Worked Example
Tide height \( h = a\sin\big(b(t - c)\big) + d \) (m) has max \( 8 \), min \( 2 \), period \( 12 \) h, first max at \( t = 3 \). Find \( a, b, c, d \).
\( a = \dfrac{8 - 2}{2} = 3 \);   \( d = \dfrac{8 + 2}{2} = 5 \);   \( b = \dfrac{2\pi}{12} = \dfrac{\pi}{6} \)
A sine peaks a quarter-period (\( 3 \) h) after \( t = c \). Max is at \( t = 3 \), so \( c = 3 - 3 = 0 \). Check: \( h(3) = 3\sin\!\big(\tfrac{\pi}{6}(3)\big) + 5 = 3(1) + 5 = 8 \) ✓
Answer: \( a = 3 \), \( b = \dfrac{\pi}{6} \), \( c = 0 \), \( d = 5 \); principal axis \( h = 5 \) m
GDC: plot \( Y1 = 3\sin(\tfrac{\pi}{6}X) + 5 \) in radian mode; G-Solv MAX confirms 8 m at \( X=3 \).
Set RADIAN mode before any SinReg / sinusoidal work — in degree mode \( b \) and the period come out as \( \dfrac{360^\circ}{b} \) and answers will not match a radian model. Find amplitude/axis from max & min; find \( b \) from the period.