Task 1:
1.
We can observe that the horizontal extent of the graph is –2 to 2, so the domain of 𝑓 is (−2,2].
The vertical extent of the graph is -∞ to +∞, so the range is (−∞,+∞)
2. The function is not one-one function because the function as it failed horizontal line test at
y=4 where it passes through more than 1 point of the graph: x1= -1, x2=1
Task 2:
(ii) Determine if E(P) is a function of P.
E(P) is a function of P because for each unique value of the output variable can be determined
from a value of the input variable. For example: P= 50000, E(P)= 40000
(iii) Find the domain and range of E(P).
We can observe that the horizontal extent of the graph is 10 to +∞, so the domain of E(P) is
[10,+∞)
, The vertical extent of the graph is 0 to+∞, so the range is [0,+∞)
(iv) Find how much export is done for 70 and 20 thousand of
production. For 70 is 60
For 20 is 10
(v) What are dependent and independent variables in this
problem? The dependent is 1 and independent is 10000
Task 3:
The graph provided shows the relationship between the weights (y in tons) of two animals and
their respective lengths (x in feet). It represents two functions:
• f: 𝑦= 𝑥Z (red curve)
• g: 𝑦=5𝑥 (blue line)
Let's address the questions:
(i) Intersection Analysis
To find the point of intersection and the rates of change in length concerning weight at that point:
1. Intersection Point: The intersection of 𝑓 and 𝑔 can be found by solving the equations 𝑦=𝑥Z
and 𝑦=5𝑥 simultaneously:
o Set 𝑥Z =5𝑥
o Solve: 𝑥Z −5𝑥=0
o 𝑥(𝑥−5)=0
This yields two solutions: 𝑥=0 and 𝑥-5=0 x=5
2. Rates of Change at Intersection Points:
o At 𝑥=0, both functions are at the origin (0,0)
o At 𝑥=5, the intersection point is (5,25)
The slopes of the tangent lines (rates of change) can be calculated using the derivatives:
• For 𝑓(𝑥)= 𝑥Z, 𝑑𝑦𝑑𝑥=2𝑥
o At 𝑥=5: 𝑑𝑦𝑑𝑥=2⋅5=10
• For 𝑔(𝑥)=5𝑥, 𝑑𝑦𝑑𝑥=5
o The slope is constant at 5 for all 𝑥
1.
We can observe that the horizontal extent of the graph is –2 to 2, so the domain of 𝑓 is (−2,2].
The vertical extent of the graph is -∞ to +∞, so the range is (−∞,+∞)
2. The function is not one-one function because the function as it failed horizontal line test at
y=4 where it passes through more than 1 point of the graph: x1= -1, x2=1
Task 2:
(ii) Determine if E(P) is a function of P.
E(P) is a function of P because for each unique value of the output variable can be determined
from a value of the input variable. For example: P= 50000, E(P)= 40000
(iii) Find the domain and range of E(P).
We can observe that the horizontal extent of the graph is 10 to +∞, so the domain of E(P) is
[10,+∞)
, The vertical extent of the graph is 0 to+∞, so the range is [0,+∞)
(iv) Find how much export is done for 70 and 20 thousand of
production. For 70 is 60
For 20 is 10
(v) What are dependent and independent variables in this
problem? The dependent is 1 and independent is 10000
Task 3:
The graph provided shows the relationship between the weights (y in tons) of two animals and
their respective lengths (x in feet). It represents two functions:
• f: 𝑦= 𝑥Z (red curve)
• g: 𝑦=5𝑥 (blue line)
Let's address the questions:
(i) Intersection Analysis
To find the point of intersection and the rates of change in length concerning weight at that point:
1. Intersection Point: The intersection of 𝑓 and 𝑔 can be found by solving the equations 𝑦=𝑥Z
and 𝑦=5𝑥 simultaneously:
o Set 𝑥Z =5𝑥
o Solve: 𝑥Z −5𝑥=0
o 𝑥(𝑥−5)=0
This yields two solutions: 𝑥=0 and 𝑥-5=0 x=5
2. Rates of Change at Intersection Points:
o At 𝑥=0, both functions are at the origin (0,0)
o At 𝑥=5, the intersection point is (5,25)
The slopes of the tangent lines (rates of change) can be calculated using the derivatives:
• For 𝑓(𝑥)= 𝑥Z, 𝑑𝑦𝑑𝑥=2𝑥
o At 𝑥=5: 𝑑𝑦𝑑𝑥=2⋅5=10
• For 𝑔(𝑥)=5𝑥, 𝑑𝑦𝑑𝑥=5
o The slope is constant at 5 for all 𝑥
