Week 8 Flashcards
Graph the following showing all steps in the transformation from the starting graph.
a) π¦=|π₯+4|β1
start with π¦=π₯+4

- Graph the following showing all steps in the transformation from the starting graph.
b) π¦=3β|π₯β2|
start with π¦=π₯β2
(b) We have been asked to draw a graph of y= 3β|xβ2|.
- Start with a graph of y=|x|(blue)
- Reflect across the x-axis: y=β|x|(green)
- Move the graph 2 to the right: y=β|xβ2|(red)
- Move the graph up 3: y=β|xβ2|+ 3 (black)

- Graph the following showing all steps in the transformation from the starting graph.
c) π¦+2+|π₯+1|=0
start with π¦=π₯+1
(c) We have been asked to draw a graph ofy+ 2 +|x+ 1|= 0.
- Start with a graph of y=|x|(blue)
- Reflect across the x-axis: y=β|x|(green)
- Move the graph 1 to the left: y=β|x+ 1|(red)
- Move the graph down 2: y=β|x+ 1|β2 (black)

- Sketch the following piecewise functions, stating the domain and range.
a) π(π₯)= {π₯β2 π₯<β1
{π₯+2 π₯β₯β1
a) π(π₯)= {π₯β2 π₯<β1
{π₯+2 π₯β₯β1
Domain: All real numbers OR ββ< x OR (ββ,β)
Range:f(x)OR (ββ,β3) βͺ [1,β)

2. Sketch the following piecewise function
a) π(π₯)= {π₯β2 x <β1
{π₯+2 π₯β₯β1
i) π(β3)
π(β3) =β3β2 =β5
2. Sketch the following piecewise function
a) π(π₯)= {π₯β2 π₯<β1
{π₯+2 π₯β₯β1
ii) π(0)
π(0) = 0 + 2 = 2
2. Sketch the following piecewise function
a) π(π₯)= {π₯β2 π₯<β1
{π₯+2 π₯β₯β1
iii) π(β1)
π(β1) =β1 + 2 = 1
2. Sketch the following piecewise functions, stating the domain and range.
b) π(π₯)={2βπ₯ π₯β€β1
{3 β1<π₯<3
{2π₯β5 π₯β₯3
b) π(π₯)={2βπ₯ π₯β€β1
{3 β1<π₯<3
{2π₯β5 π₯β₯3
Domain: All real numbers OR ββ< x OR (ββ,β)
Range:g(x)β₯1 OR [1,β)

2. Sketch the following piecewise function
b) π(π₯)={2βπ₯ π₯β€β1
{3 β1<π₯<3
{2π₯β5 π₯β₯3
i) π(β4)
b) π(π₯)={2βπ₯ π₯β€β1
{3 β1<π₯<3
{2π₯β5 π₯β₯3
g(β4) = 2ββ4 = 6
π¦=|π₯|
with a vertical line on either side of the π₯.
|π₯|can be thought of as how far away from zero π₯ is, where direction is not important.
So no negatives. Hence |3|=3and |β3|=3as both +3and β3are both 3units away from the origin.
Graphing Absolute Functions
Looking at both graphs, where π₯ is positive (on the RHS of the y-axis), the graphs are
identical. But where π₯ is negative, (on LHS of π¦-axis) then π¦ = |π₯| is a mirror image of π¦ =
π₯ (reflected in the π₯-axis)

Domain and Range of π¦ = |π₯|
Clearly from the graph of π¦ = |π₯|
Domain: -β < π₯ < β
Range: 0 β€ π¦ < β

Graph π¦ = |π₯ - 3| + 2

Sketch π¦ = -1 - |3π₯ + 2|

Hybrid or Piecewise Functions
Suppose you are recording some growth in population of some cells over time.
From π‘ = 0 to π‘ = 2 (hours)
the number of cells (in 100βs)
can be modelled by πΆ(π‘) = π‘2 + 1.
So at π‘ =0 we have
πΆ(0) = 02 + 1 = 1 (or 100) cells.
You observe that after π‘ = 2 the population is modelled by the equation πΆ(π‘) = 3π‘ - 1 up until π‘ = 6
When we observe two or more functions over a set domain, we call it a hybrid or piecewise
function. Our example above can be written.





finding the Inverse Function
π(π₯) = _1/2_π₯ + 5

Sketch
y=|x+ 2|

sketch
y=|x|β4

sketch
y=|x+ 3|β1

sketch
y=β|x|+ 2

sketch
y=β|xβ1|+ 3

eval
f(-1)

f(β1)
=β(β1) + 4
= 1 + 4
= 5
eval
f(0)

f(0)
=β0 + 4
= 4
eval
f(1)

f(1)
= 2Γ1β1
= 2β1
= 1
eval
f(2)

f(2)
= 2Γ2β1
= 4β1
= 3
Sketch f(x)


Determine the domain and range of f(x)

Domain: All (real) values of x
Range:f(x)β₯1
eval
g(-3)

g(β3)
= 2β(β3)
= 2 + 3
= 5
eval
g(-2)

g(β2)
= 2β(β2)
= 2 + 2
= 4
eval
g(0)

g(0) = 3
eval
g(3)

g(3)
= 2Γ3β4
= 6β4
= 2
Sketch g(x)


Determine the domain and range of g(x).

Domain: All (real) values of x
Range: g(x)β₯2
find its inverse function
f(x) = 7xβ2

find its inverse function
g(x) = 5e3x

find its inverse function
f(x) =4x+3/5β1

find its inverse function.
g(x) = 3/xβ2

find its inverse function.
h(x) = 3β2exβ1

find its inverse function
j(x) = 2 + 5 ln(x)

Show that the inverse of
f(x) = x+ 1/xβ1
is itself.

Given the functions f(x) = 5x and g(x) = 2 cos(x), find the following composite functions.
f(g(x)) (This could also be written as fβ¦g)
f(x) = 5x and g(x) = 2 cos(x)
f(g(x)) = f(2 cos(x))
5Γ(2 cos(x))
= 10 cos(x)
Given the functions f(x) = 5x and g(x) = 2 cos(x), find the following composite functions.
g(f(x)) (This could also be written as gβ¦f)
g(f(x)) = g(5x)
= 2 cos(5x)
Given the functions g(x) = 4βx2 and h(x) = 3xβ2, find the following composite functions.
g(h(x)) (This could also be written as gβ¦h)
g(x) = 4βx2 and h(x) = 3xβ2
g(h(x)) = g(3xβ2)
= 4β(3xβ2)2
= 4β(9x2β12x+ 4)
= 4β9x2+ 12xβ4
=β9x2+ 12x
Given the functions g(x) = 4βx2 and h(x) = 3xβ2, find the following composite function
h(g(x)) (This could also be written as hβ¦g)
h(g(x)) = h(4βx2)
= 3(4βx2)β2
= 12β3x2β2
= β3x2+ 10
Given the functions
f(x) = 4x+ 2,
g(x) = 1β3x
and h(x) = 2x2+ 1,
find the composite function fβ¦gβ¦h.



shown is old graph*


State the (natural) domain and range for the function y=ex
make a table if it helps
Domain: All real values of x OR ββ< x <β.
Range: All positive values of y OR 0< y <β.
Note that y= 0 is NOT included in the range.