Question 1:
a) (2 marks) True or false? 𝒉(𝒙) = 𝒙𝟒 + 𝒙 is a one-to-one function. If you believe the
statement is true, prove it using the definition of a one-to-one function. If you believe it is
false explain why.
To prove the function is not injective, it will first be factorised into a more friendly form:
ℎ(𝑥) = 𝑥 4 + 𝑥 = 𝑥(𝑥 3 + 1)
Using the above factorized form, it is easy to solve for the roots of the polynomial:
0 = 𝑥(𝑥 3 + 1)
⇒ 𝑥 = 0,
𝑥3 + 1 = 0
∴ 𝑥 = 0,
𝑥 = −1
Two distinct values of 𝑥1 = 0 and 𝑥2 = −1 provide the same value of 𝑓(𝑥1 ) = 𝑓(𝑥2 ) = 0.
However, if the mapping ℎ is claimed as being ‘one-to-one’ or ‘injective’, it must map a unique value
𝑓(𝑥1 ) from the domain (say, 𝑋), to a unique value 𝑥1 in the range (say, 𝑌). Yet, as solving for the root(s)
of the function demonstrates, two unique values from the function’s domain 𝑋 (being 𝑥1 = 0 and
𝑥2 = −1), map to a single value in range 𝑌 (being 𝑓(𝑥1 ) = 𝑓(𝑥2 ) = 0). As such, ℎ(𝑥) = 𝑥 4 + 𝑥 cannot
be considered a ‘one-to-one’ function.
b) (2 marks) True or False? If 𝒇 is a function defined on the domain of all real numbers and has
an inverse, 𝒇−𝟏 , then 𝒇−𝟏 is defined on the domain of all real numbers. If true, explain why,
if false, provide a specific function which disproves the statement.
Let us define a mapping 𝑓: ℝ → ℝ+ , 𝑓(𝑥) = 𝑥 2 . We can solve for 𝑓 −1 as follows:
𝑓(𝑥) = 𝑥 2 = 𝑦
⇒ 𝑥 = ±√𝑦
∴ 𝑓 −1 (𝑥) = ±√𝑥
The inverse function 𝑓 −1 cannot yield a value outside of the domain of the original function 𝑓, as the
range of the inverse is the domain of the original function: range(𝑓 −1 ) = dom(𝑓), and by extension,
range(𝑓) = dom(𝑓 −1 ). The function 𝑓(𝑥) = 𝑥 2 is defined for all real numbers, and its range includes
all positive real numbers and zero. As such, the inverse function 𝑓 −1 (𝑥) = ±√𝑥 has a domain across
all positive real numbers and zero; and its range includes all the real numbers. However, this statement
is in direct contradiction of the initial claim that “If 𝑓 is a function defined on the domain of all real
numbers and has an inverse, 𝑓 −1 , then 𝑓 −1 is defined on the domain of all real numbers”, which means
the statement must be false.
To further emphasise this point, for the set of real numbers, the square root of a negative number is
not defined, while the square of a negative number is defined. Similarly, the geometric interpretation
of the inverse 𝑓 −1 (which reflects the graph of 𝑓 along the symmetry line 𝑦 = 𝑥) generates a sideways
parabola undefined for negative real numbers, despite the fact that the domain of 𝑓 exists for the set
of real numbers. This further shows that the statement is false.
𝒙−𝟏
c) (3 marks) Let 𝒇(𝒙) = 𝒙−𝟐 and 𝒈(𝒙) = 𝐥𝐧(𝒙). What is the domain and range of 𝒈 ∘ 𝒇?
𝑥−1
As 𝑓(𝑥) = 𝑥−2 and 𝑔(𝑥) = ln(𝑥):
𝑥−1
𝑥−1
𝑔 ∘ 𝑓 = 𝑔(𝑓(𝑥))
𝑥−1
𝑔 ∘ 𝑓 = ln (
)
𝑥−2
If ln (𝑥−2) = ln(𝑢), where 𝑢 = 𝑥−2, the function is undefined for 𝑢 outside of the interval (0, ∞). As
such, solving for 𝑢 ≤ 0 will yield all values outside of the function’s domain:
𝑥−1
=0
𝑥−2
⇒ 𝑥 − 1 = 0,
∴ 𝑥 = 1,
𝑥−2=0
𝑥=2
ln(𝑢) must be undefined at 𝑥 = 1, 𝑥 = 2. To determine what other points are undefined, the nature
of the inequality will be examined for the intervals (−∞, 1), (1, 2), (2, ∞) by testing a value from each
𝑥−1
interval to determine where 𝑥−2 is positive and/or negative within that interval, and by extension,
determine whether ln(𝑢) is defined over that interval.
Interval: (−∞, 1)
Interval: (1, 2)
Interval: (2, ∞)
(0) − 1 −1 1
=
=
(0) − 2 −2 2
3
1
(2) − 1
= 2 = −1
1
3
( ) − 2 −2
2
5
3
(2) − 1
= 2=3
1
5
( )−2 2
2
3
Testing 𝑥 = 0
Testing 𝑥 = 2
As 𝑢 > 0, ln(𝑢) is defined
𝑥−1
As 𝑢 < 0, ln(𝑢) is undefined
5
Testing 𝑥 = 2
As 𝑢 > 0, ln(𝑢) is defined
Thus, the domain of ln (𝑥−2) is 𝑥 ∈ {(−∞, 1), (2, ∞)} (from negative infinity up to, but not
including, 1; and from positive infinity down to, but not including, 2).
𝑥−1
The range of ln (𝑥−2) contains all real numbers excluding zero, as the base function ln(𝑥) has not
been transformed through dilations, reflections or translations of any kind, and the exception of zero
𝑥−1
occurs because the value of 𝑥−2 only approaches 1 as 𝑥 approaches positive and negative infinity.
Additionally, 0 < 𝑢 < 1 when 𝑥 < 1 (meaning the function is negative), while 𝑢 > 1 when 𝑥 > 2
(meaning the function is positive). As a consequence of this, the function has vertical asymptotes at
𝑥 = 1 (where ln(𝑢) tends towards positive infinity) and 𝑥 = 2 (where ln(𝑢) tends towards negative
infinity), as well as a horizontal asymptote at 𝑦 = 0, which is reached as 𝑥 → ∞ and 𝑥 → −∞.
Thus, the graphed logarithm appears hyperbolic.
d) (3 marks) Use the limit laws and one of the useful sequences to remember, in sections 3.4
and 3.5 of the workbook, to calculate the limit of the sequence
𝟏 𝟐𝒏−𝟏
𝒂𝒏 = (− )
𝟐
Before taking the limit of the sequence, it will be tested for initial values of 𝑛 to check if an underlying
pattern exists:
1 2(1)−1
1 1
1
𝑎1 = (− )
= (− ) = −
2
2
2
2(2)−1
3
1
1
1
𝑎2 = (− )
= (− ) = −
2
2
8
1 2(3)−1
1 5
1
𝑎3 = (− )
= (− ) = −
2
2
32
𝑐
By testing values of 𝑎𝑛 , it seems that the limit will converge to − ∞ = 0 as 𝑛 → ∞.
However, this can only be verified by solving the limit of the function using limit laws:
1 2𝑛−1
lim 𝑎𝑛 = lim (− )
𝑛→∞
𝑛→∞
2
= lim (−
𝑛→∞
12𝑛−1
)
22𝑛−1
( lim 12𝑛−1 )
= ( lim −1) 𝑛→∞
𝑛→∞
( lim 22𝑛−1 )
𝑛→∞
( lim 12𝑛 ) ( lim 1−1 )
𝑛→∞
= ( lim −1) 𝑛→∞
𝑛→∞
2𝑛
( lim 2 ) ( lim 2−1 )
𝑛→∞
𝑛→∞
( lim (12 )𝑛 ) ( lim 21 )
𝑛→∞
= ( lim −1) 𝑛→∞
𝑛→∞
2
𝑛
( lim (2 ) ) ( lim 11 )
𝑛→∞
𝑛→∞
( lim 1𝑛 ) ( lim 2)
𝑛→∞
= ( lim −1) 𝑛→∞
𝑛→∞
( lim 4𝑛 ) ( lim 1)
𝑛→∞
𝑛→∞
If 𝑐 is a constant, lim 𝑐 = 𝑐 as 𝑛 does not impact the value of 𝑐. Using this statement, the above
𝑛→∞
expression is simplified for constant values:
As lim 1𝑛 = 1:
𝑛→∞
( lim 1𝑛 ) (2)
= (−1) 𝑛→∞
( lim 4𝑛 ) (1)
𝑛→∞
= (−1)
(1)(2)
( lim 4𝑛 ) (1)
𝑛→∞
=
−2
( lim 4𝑛 )
𝑛→∞
As lim 𝑐 𝑛 is divergent for |𝑐| > 1, 𝑐 = 4 will cause the denominator to diverge, approaching infinity:
𝑛→∞
=
−2
∞
As the numerator is a constant term, the overall effect is that the fraction will tend towards zero:
1 2𝑛−1
Thus, for the sequence 𝑎𝑛 = (− 2)
=0
, lim 𝑎𝑛 = 0.
𝑛→∞
Question 2:
(10 marks) Inside a 1,728 km2 Queensland forest several companies have been chopping
down trees to build houses. The government has decided to create a national park within
the forest, which cannot be chopped down, to protect several native species of insects. To
aid the government’s efforts, entomologists have provided the government with a “SpeciesArea Relationship,” a function, 𝑺, that outputs the expected number of insect species in a
patch of forest, given the area of that patch, 𝒂, in km2,
𝑺(𝒂) = 𝟒𝟎𝟎𝟎𝒂𝟏/𝟑
a) (2 marks) What is the domain and range of 𝑺? Explain their meaning biologically.
As area cannot be negative, 𝑎 ≥ 0 is the lower bound of the function’s domain. As the forest has a
maximum size of 𝑎 = 1728 km2 , it follows that the maximum area which can be input is 𝑎 = 1728.
As such, the domain of 𝑆(𝑎) is 𝑎 ∈ [0, 1728]; dom(𝑆(𝑎)) = [0, 1728].
To determine the range, it was first noted that the cube root of 𝑥 increases as 𝑥 increases, and also
that the cube root is a one-to-one function. As such, the maximum and minimum values of the
function will be located at the maximum and minimum values of the domain:
𝑆(0) = 4000(0)1/3 = 0
𝑆(1728) = 4000(1728)1/3 = 4000(12) = 48,000
Thus, range(𝑆(𝑎)) = [0, 48000]
Biologically speaking, this means that a forest of size 1728km2 is not estimated to exceed a value of
48,000 unique insect species, and that a forest of size 0km2 is not estimated to have any insect species.
b) (3 marks) Use Matlab to graph 𝑺(𝒂) versus 𝒂. Label the 𝒙 and the 𝒚 axes, and add a title to
your graph. Submit your code and a printout of the graph. Refer to the Matlab activity sheet
on functions for help with plotting graphs in Matlab.
a = linspace(0,1786,1787);
S = 4000 * a.^(1/3);
plot(a,S)
legend('Insect species S(a)')
ylabel({'Insect species; S(a)'});
xlabel({'Area; a'});
title({'Estimated insect species as a function of area'});
c) (3 marks) Find the inverse of 𝑺 and explain what it means.
If 𝑆(𝑎) = 4000𝑎1/3, let 𝑆 = 4000𝑎1/3 and solve for 𝑎 in order to determine 𝑆 −1 (𝑎):
1
𝑆 = 4000𝑎3
1
𝑆
= 𝑎3
4000
(
Thus,
𝑆 3
) =𝑎
4000
𝑎 3
𝑆 −1 (𝑎) = (
)
4000
However, as the variables were swapped to determine the inverse function, this means that in the
above expression, 𝑆 refers to the area, while 𝑎 refers to the species present, as the number of species
must be used as a function input for 𝑆 −1 (𝑎) in order to determine the inhabited area by the species.
d) (2 marks) How big must the new national park be if the government wants it to protect
20,000 species?
Let 𝑆(𝑎) = 20000, and hence solve for 𝑎:
𝑆(𝑎) = 4000𝑎1/3
20000 = 4000𝑎1/3
20000
= 𝑎1/3
4000
5 = 𝑎1/3
125 = 𝑎
The estimated necessary area for 20000 unique species of insects to be protected is 125km2. However,
as the equation serves as an estimate, it is recommended that the actual preserved area is larger than
this value, e.g. 130km2.