🔬 Tutorial problems gamma \(\gamma\)#
Note
This problems are designed to help you practice the concepts covered in the lectures. Not all problems may be covered in the tutorial, those left out are for additional practice on your own.
\(\gamma\).1#
Suppose that the cost of producing \(Q\) units of a commodity is given by \(C(Q) = 1, 000 + 300Q + Q^2\).
Compute \(C(0)\), \(C(100)\), and \(C(101) - C(100)\).
Compute \(C(Q + 1) - C(Q)\) and explain the meaning of this expression.
[Sydsæter, Hammond, Strøm, and Carvajal, 2016] Exercises for Section 4.2, Question 6
Suppose that the cost of producing \(Q\) units of a commodity is given by \(C(Q)=1,000+300Q+Q2\). We have
Note that \(C(Q+1)-C(Q)\) is the difference between the total cost of producing \(Q+1\) units of a commodity and the total cost of producing \(Q\) units of that commodity. As such, it is the incremental cost that is incurred by producing the \((Q + 1)\)-th unit of the commodity. Since the incremental increase is a one unit increase in this case, this incremental cost is known as the marginal cost of the \((Q + 1)\)-th unit of the commodity.
\(\gamma\).2#
Consider a rule of association that assigns to each person their sibling. Is this rule a function? Why or why not?
To properly answer the question it is important start with the definition of domain and co-domain of the described rule of association. Properly stated problem would have defined both explicitly, but if not, the answer should include several possible cases.
According to the definition a function is a mapping that associates each element of a domain with uniquely determined element of a co-domain.
If the domain is all people, then the described correspondence can not be a function because there are people who do not have siblings, and therefore are not associated to any element of the co-domain.
If we restrict the domain to only people with siblings, there is still a problem because there are people who have more than one sibling, and therefore are associated with more than one element of the co-domain.
If we further restrict the domain to only people with precisely one sibling. In this case, the association is a function. The co-domain in this case should include at least all people with at a sibling.
\(\gamma\).3#
Consider the function \(f(x) = \frac{3x+6}{x-2}\).
Find the domain of this function
Show that \(5\) belongs to the range of this function.
Show that \(3\) does not belong to the range of this function.
[Sydsæter, Hammond, Strøm, and Carvajal, 2016] Exercises for Section 4.2, Question 14
To show that a particular function value is in range, find a value for \(x\) corresponding to it and observe it is in the domain of the function.
Note that \(f(x)\) will be defined whenever the denominator \((x-2)\) does not equal zero.Thus the only restriction on the domain of this function is that \(x \ne 2\). Therefore the domain for this function is the set \(D = \{x \in \mathbb{R} : x \ne 2\} = \mathbb{R} \setminus \{2\}\).
Suppose that \(f(x) = 5\). This requires that \(f^{-1}(5)\) is defined:
Thus, for \(x=8\) \(f(x) = 5\) and therefore \(5\) belongs to the range of \(f\).
Suppose instead that f(x) = 3. This requires that
This equation does not have any solutions. Therefore, \(f^{-1}(3)\) is undefined, and so \(3\) does not belong to the range of \(f\).
\(\gamma\).4#
If \(f(x)=3x-x^3\) and \(g(x)=x^3\), compute the following:
\((f+g)(x)\)
\((f-g)(x)\)
\((fg)(x)\)
\((f/g)(x)\)
\(f(g(1))\)
\(g(f(1))\)
\((f \circ g)(x)\)
\((g \circ f)(x)\)
[Sydsæter, Hammond, Strøm, and Carvajal, 2016] Exercises for Section 5.2, Question 3
For standard arithmetic operations \(\bullet\) notation \((f \bullet g)(x)\) is equivalent to \(f(x) \bullet g(x)\) point by point for all \(x\) in the intersection of the domains for \(f(x)\) and \(g(x)\).
\((f+g)(x) = 3x\)
\((f-g)(x) = 3x - 2x^3\)
\((fg)(x) = 3x^4 - x^6\)
\((f/g)(x) = \frac{3}{x^3} - 1\)
\(f(g(1))=f(1)=2\)
\(g(f(1))=g(2)=8\)
\((f \circ g)(x) = 3(x^3) - (x^3)^3 = 3x^3 - x^9\)
\((g \circ f)(x) = (3x - x^3)^3 = 27x^3 - 27x^5 + 9x^7 - x^9\)
\(\gamma\).5#
Let \(f(x)=3x+7\). Compute \(f(f(x))\), and find the value \(x^\star\) such that \(f(f(x^\star))=100\).
[Sydsæter, Hammond, Strøm, and Carvajal, 2016] Exercises for Section 5.2, Question 3
\(f(f(x)) = f(3x+7)=3(3x+7)+7=9x+28\). The equality \(f(f(x^\star))=100\) implies \(9x^\star+28=100\), so \(x^\star=8\).
\(\gamma\).6#
Find the domains of the following functions and then illustrate those domains in the \((x, y)\)–coordinate-plane.
\(f(x,y)= \frac{x^2+y^3}{y-x+2}\)
\(f(x,y)=\sqrt{2-(x^2 +y^2)}\)
\(f(x,y)=\sqrt{(4-x^2 -y^2)(x^2 +y^2 -1)}\)
[Sydsæter, Hammond, Strøm, and Carvajal, 2016] Exercises for Section 11.1, Question 6
Disregard the fact that the questions asks about function of two variables and focus on finding the domain of these functions in the form of a restriction on \((x,y)\).
Part 1.
Consider the function \(f(x,y) = \frac{x^2+y^3}{y-x+2}\). The only restriction on the domain of this function is the impossibility of division by zero. Thus the only restriction on the domain of this function is that \(y - x + 2 \ne 0\). In other words, the only restriction on the domain of this function is that it not include any points on the line \(y = x - 2\). Thus the domain of this function is
This is the entire two-dimensional Euclidean coordinate-plane excluding the points on the line \(y = x - 2\).
Note
Because the question explicitly asks to illustrate, full marks for the question like this will be awarded for the answer that includes both the derivation of this result and a sketch of the domain in a two-dimensional coordinate plane.
Part 2.
Consider the function \(f(x,y)=\sqrt{2-(x^2 +y^2)}\). Restriction on the domain of this function stems from the fact that the function \(\sqrt{x}\) is defined only for non-negative \(x\). Thus the only restriction on the domain of this function is that \(2 - (x^2 + y^2) \ge 0\). This requires that \(x^2 + y^2 \le 2\). As such, the domain for this function is the set
This region consists of both the boundary of, and area inside, a circle of radius \(\sqrt{2}\) that is centred on the origin, that is, the point \((0, 0)\).
Part 3.
Consider the function \(f(x,y)=\sqrt{(4-x^2 -y^2)(x^2 +y^2 -1)}\). As in the previous part, the only restriction on the domain of this function stems from the fact that the function \(\sqrt{x}\) is defined only for non-negative \(x\).
This might occur in any of the following ways.
Both \(4-x^2 -y^2 >0\) and \(x^2 +y^2 -1 >0\), or
Both \(4-x^2 -y^2 <0\) and \(x^2 +y^2 -1<0\), or
\(4 - x^2 - y^2\) = 0, or
\(x^2 +y^2 -1=0\).
The first of these cases requires that \(1 < x^2 + y^2 < 4\). This is the area between the circle centred on the origin with radius one and the circle centred on the origin with radius two, not including either of the boundaries.
The second of these cases requires that \(4 < x^2 + y^2 < 1\), which is impossible because \(4>1\). Thus we know that this case cannot occur.
The third of these cases requires that \(x^2 + y^2 = 4\). This is the set of points on the boundary of a circle of radius two that is centred on the origin.
The third of these cases requires that \(x^2 + y^2 = 1\). This is the set of points on the boundary of a circle of radius one that is centred on the origin.
After combining cases one, three, and four, we conclude that the domain for this function is
This is the area between the circle centred on the origin with radius one and the circle centred on the origin with radius two, including both of the boundaries. It looks like a dough-nut or a simple washer when viewed from side-on.