Exercises 2

This page allows you to practice some exercises on Functions of One Variable. If you notice that you have difficulties, we advise you to go over the corresponding sections in the book.

These questions correspond to the following sections: - Section 4.4 - Linear Functions - Section 4.5 - Linear Models - Section 4.6 - Quadratic Functions

Question 7

Find lines of the form $y = a+bx$ passing through the following pairs of points $(x,y)$: (a) $(2,3)$ and $(5,8)$

(b) $(-1, -3)$ and $(2, -5)$

(c) $\left(\frac{1}{2}, \frac{3}{2} \right)$ and $\left(\frac{1}{3}, -\frac{1}{5}\right)$

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(a) When $x$ goes from $2$ to $5$ (i.e., $\Delta x = 5-2 =3$) then $y$ from $3$ to $8$ (i.e., $\Delta y = 8-3=5$). Thus, for each one-unit increase in $x$, we have that $y$ increases by $b = \frac{\Delta y}{\Delta x} = \frac{5}{3} = 1 \frac{2}{3}$. Given this value for $b$, we can evaluate $a + bx$ at $x=2$. From the first coordinate, we know that $y=3$ at that point. Thus, we have $a + 1 \frac{2}{3} \cdot 2=3 \Leftrightarrow a = 3 - 3 \frac{1}{3} = - \frac{1}{3}$. In other words the intercept (i.e., the value for $y$ when $x=0$ for the straight line passing through the two coordinates provided in the question) is $-\frac{1}{3}$. We can write this compactly as $y = -\frac{1}{3} + 1 \frac{2}{3} x$.

(b) Similarly to Question 7a, we have that $b=\frac{\Delta y}{ \Delta x} = \frac{-2}{3} = - \frac{2}{3}$. Solving for $a$ using the first coordinate, we have $a-\frac{2}{3}\cdot(-1)=-3 \Leftrightarrow a = -3 \frac{2}{3}$. Thus, $y=-3\frac{2}{3} - \frac{2}{3} x$.

(c) We have $b=\frac{\Delta y}{\Delta x}=\frac{-\frac{1}{5} - \frac{3}{2}}{\frac{1}{3}-\frac{1}{2}}=\frac{-\frac{2}{10} - \frac{15}{10}}{\frac{2}{6}-\frac{3}{6}}=\frac{-\frac{17}{10}}{-\frac{1}{6}}={\frac{17 \cdot 6 }{10}}=\frac{17 \cdot 3}{5} = 10 \frac{1}{5}$. Solving for $a$ using the second coordinate, we have $a + \frac{17 \cdot 3}{5} \cdot \frac{1}{3} = - \frac{1}{5} \Leftrightarrow a = -\frac{1}{5} - \frac{17}{5} = -\frac{18}{5} = -3 \frac{3}{5}$. Thus, $y=-3\frac{3}{5} + 10 \frac{1}{5} x$.

Question 8

Decide which of the following relationships between two variables are linear. (a) $5y + 2x = 2$

(b) $P = 10(1-0.3t)$

(c) $C = \left( \frac{1}{2} x + 2 \right)(x-3)$

(d) $p_{1}x_{1} + p_{2}x_{2} = R$, where $p_{1}, p_{2},$ and $R$ are constants

Show answer

(a) Linear: we can subtract $2x$ from both sides, and divide both sides of the new equation by $5$ to find an expression for $y$ as a linear function of $x$.

(b) Linear: we only have to expand the brackets to find that $P$ s a linear function of $t$.

(c) Quadratic (so not linear): by expanding the parentheses, we find $C=0.5 x^2 + \frac{1}{2} x - 6$.

(d) Linear: we can subtract $p_1 x_1$ from both sides of the equation. Dividing the subsequent equation by $p_2$, we have expressed $x_2$ as linear function of $x_1$.

Question 9

A printing company quotes the price of €1400 for producing 100 copies of a report, and €3000 for 500 copies. Assuming a linear relation, what would be the price of printing 300 copies?

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Let $x$ denote the number of copies and $y$ the price in euros. Effectively, we are given coordinate $(x,y)=(100,1400)$, coordinate $(x,y)=(500,3000)$, and the assumption of a linear relation between $x$ and $y$ (i.e., we can write $y$ in the form $a+bx$ for constants $a$ and $b$).

Now, we can use the same approach as before:

• Slope $b = \frac{\Delta y}{\Delta x} = \frac{3000-1400}{500-100} = \frac{1600}{400} = 4$ euros per copy. So for each additional copy, you pay an additional 4 euros.

• Find intercept $a$ by solving equation for one coordinate under linear relation, given $b=4$. Using the first coordinate: $1400=a+4 \cdot 100 \Leftrightarrow a=1000$ euros.

Thus, $y=1000+4x$.

Question 10

Determine the equations for the following straight lines:

(a) $\ell_{1}$ passes through $(1,3)$ and has a slope of $2$.

(b) $\ell_{2}$ passes through $(-2, 2)$ and $(3,3)$.

Show answer

(a) We write $y=a+bx$. The slope is given: $b=2$. Solving for $a$ using the provided coordinate: $3 = a + 2 \cdot 1 \Leftrightarrow a = 1$. Thus, $y=1+2x$.

(b) We write $y=a+bx$. We have $b=\frac{\Delta y}{\Delta x} = \frac{3-2}{3-(-2)} = \frac{1}{5}$. Solving for $a$ using the second coordinate that is provided: $3=a + \frac{1}{5} \cdot 3 \Leftrightarrow a = 3 - \frac{3}{5} = 2 \frac{2}{5}$. Thus, $y= 2 \frac{2}{5} + \frac{1}{5}x$.

Question 11

Find the equilibrium price for each of the following linear models of supply and demand:

(a) $D=75-3P$ and $S=2P$

(b) $D = 100 - \frac{1}{2}P$ and $S = -20 + \frac{1}{2}P$

Show answer

(a) $D=S \Leftrightarrow 75 - 3P = 2P \Leftrightarrow 75 = 5P \Leftrightarrow P = \frac{75}{5}= 15$

(b) $D=S \Leftrightarrow 100 - \frac{1}{2}P = -20 + \frac{1}{2} P \Leftrightarrow 100 = -20 + P \Leftrightarrow P = 120$

Question 12

The total cost $C$ of producing $x$ units of some commodity is a linear function of $x$. Records show that on one occasion, 100 units were made at a total cost of €200, and on another occasion, 150 units were made at a total cost of €275. Express the linear equation for total cost in terms of the number of units produced.

Show answer

Let $y$ denote the price in euros. We assume a linear relation between $x$ and $y$. That is, we can write $y=a+bx$. We are effectively given the following two coordinates $(x,y)=(100,200)$ and $(x,y)=(150,275)$.

Hence, $b=\frac{\Delta y}{\Delta x} = \frac{275-200}{150-100} = \frac{75}{50} = 1 \frac{1}{2}$ euros per unit. Solving for $a$ using the first coordinate: $200=a+ 1 \frac{1}{2} \cdot 100 \Leftrightarrow 200 = a + 150 \Leftrightarrow a = 50$. Thus: $y=50+ 1 \frac{1}{2} x$.

Question 13

The expenditure of a household on consumer goods, $C$, is related to the household's income, $y$, in the following way. When the household's income is €1000, the expenditure on consumer goods is €900, and whenever income increases by €100, the expenditure on consumer goods increases by €80. Express the expenditure on consumer goods as a function of income, assuming a linear relationship.

Show answer

We seek to express $C$ in the following form: $C=a + by$ for appropriate constants $a$ and $b$. Recognising that $y$ now is the value of the argument and $C$ the value of the function, we have that $b=\frac{\Delta C}{\Delta y} = \frac{80}{100} = \frac{4}{5}$. Notice that we are also provided with a single coordinate: $(y,C)=(1000,900)$. Using this coordinate to solve for $a$: $900=a + \frac{4}{5} \cdot 1000 \Leftrightarrow 900 = a + 800 \Leftrightarrow a = 100$ euros. Hence, $C=100+ \frac{4}{5} y$.

Question 14

Determine the maximum/minimum points of the following functions. Moreover, for each function, find all the roots (i.e., points where the function equals zero) and where possible write each function in the form $a(x - x_{1})(x-x_{2})$ where $x_1$ and $x_2$ are integers.

(a) $x^2 + 4x$

(b) $x^2 + 6x + 18$

(c) $-3x^2 + 30x - 30$

(d) $9x^2 - 6x - 44$

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Recap: a quadratic function of the form $y=ax^2 + bx + c$, for constants $a$, $b$, and $c$ is a parabola opening upwards (Dutch: dalparabool) if $a>0$ and a parabola opening downwards (Dutch: bergparabool) if $a<0$.

For a parabola, its extreme value is found at $x^* =- \frac{b}{2a}$. In case of a parabola opening upwards, this is the global minimum point. In case of a parabola opening downwards, this is the global maximum point.

If we can write the quadratic function in the form $a(x - x_1) (x - x_2)$ its roots are given by $x_1$ and $x_2$. In case it is (too) difficult to write the quadratic function in that form, you can instead use the quadratic formula (Dutch: $abc$-formule), which states $x_0 = \frac{-b \pm \sqrt{d}}{2a}$ are the roots, where $d=b^2 - 4ac$ (aka the discriminant). In case $d<0$, the quadratic function has no roots. In case $d=0$, the quadratic function equals zero at precisely one point. In case $d>0$, the quadratic function has two roots.

(a) Here, $a=1$, $b=4$, $c=0$. Because $a>0$ it is a parabola opening upwards. Hence, $x^* = -\frac{b}{2a}= -2$ is the global minimum point. We can write this function in the form $x(x+4)$. Thus, its roots are given by $x=0$ and $x=-4$

(b) Here, $a=1$, $b=6$, $c=18$. Because $a>0$ it is a parabola opening upwards. Thus,$x=-\frac{b}{2a}=-3$ is the global minimum point. By applying the quadratic formula, we find $x_0 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-6 \pm \sqrt{36-4\cdot 1 \cdot 18 }}{2} = \textrm{NaN}$, because the discriminant is negative. In other words, the function has no roots, and cannot be written in the form $a(x-x_1)(x-x_2)$.

(c) Here, $a=-3$, $b=30$, $c=-30$. Because $a<0$ it is a parabola opening downwards. Thus, $x= - \frac{b}{2a} = 5$ is the global maximum point. By applying the quadratic formulate, we find $x_0 = \frac{-30 \pm \sqrt{900-4\cdot (-3) \cdot (-30)}}{-6}=5 \pm \frac{1}{6}\sqrt{540} = 5 \pm \frac{1}{6} \sqrt{ 2^2 \cdot 3^3 \cdot 5} = 5 \pm \sqrt{15}$.

(d) Here, $a=9$, $b=-6$, $c=-44$. Because $a>0$ it is a parabola opening upwards. Thus, $x=-\frac{b}{2a}=\frac{1}{3}$ is the global minimum point. By applying the quadratic formula, we find the following roots: $x_0 = \frac{6 \pm \sqrt{36 - 4 \cdot 9 \cdot (-44)}}{18} = \frac{1}{3} \pm \frac{1}{18} \sqrt{36 + 36 \cdot 44 } = \frac{1}{3} \pm \frac{1}{18} \sqrt{36 \cdot 45} = \frac{1}{3} \pm \frac{1}{18} \sqrt{6 \cdot 6 \cdot 3 \cdot 3 \cdot 5} = \frac{1}{3} \pm \sqrt{5}$

Question 15

Find solutions to the following equations, where $a$ and $b$ are parameters.

(a) $x^2 - 3ax + 2a^2 = 0$

(b) $x^2 - (a+b)x + ab = 0$

(c) $2x^2 + (4b - a)x = 2ab$

Show answer

(a) We can write this equation in the form $(x-2a)(x-a)=0$, which implies $x-2a=0$ or $x-a=0$. Thus, $x=2a \lor x=a$.

(b) We can write this equation in the form $(x-a)(x-b)=0$. Thus, $x=a \lor x=b$.

(c) First notice that we can rearrange this equation as follows: $2x^2 + (4b-a)x - 2ab = 0$. Now, we can write the equation as $(2x-a)(x+2b)=0$. Thus, $2x-a =0 \Leftrightarrow x = \frac{1}{2}a$ or $x+2b=0 \Leftrightarrow x=-2b$.

Notice: finding such factorisations may take quite some practice to get right. Do not be disheartened if you are unable to find the right solution right away! Just take your time to practise. Please keep in mind that factorisations (not just for quadratic equations and formulas) play a big role in mathematics, statistics, linear algebra, and so on.

Question 16

If a cocoa shipping firm sells $Q$ tons of cocoa in the UK, the price it receives is given by $P_{U} = \alpha_{1} - \frac{1}{3}Q$. On the other hand, if it buys $Q$ tons of cocoa from its only source in Ghana, the price it has to pay is given by $P_{G} = \alpha_{2} + \frac{1}{6}Q$. In addition, it costs $\gamma$ per ton to ship cocoa from its supplier in Ghana to its customers in the UK (its only market). The numbers $\alpha_{1}, \alpha_{2},$ and $\gamma$ are all positive.

(a) Express the cocoa shipper’s profit as a function of $Q$, the number of tons shipped.

(b) Assuming that $\alpha_1 − \alpha_2 − \gamma > 0$, find the profit-maximizing shipment of cocoa. What happens if $\alpha_{1} − \alpha_{2} − \gamma \leq 0$?

(c) Suppose the government of Ghana imposes an export tax on cocoa of $\tau$ per ton. Find the new expression for the shipper’s profits and the new profit-maximizing quantity shipped.

(d) Calculate the Ghanaian government’s export tax revenue as a function of $\tau$.

(e) Advise the Ghanaian government on how to obtain as much tax revenue as possible.

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(a) Total costs are given by the sum of shipping costs and buying costs. The shipping costs are given by $\gamma Q$. The buying costs are given by buying price (i.e., $\alpha_2 + \frac{1}{6} Q$) times the quantity bought (i.e., $Q$). Hence, buying costs equal $\alpha_2 Q + \frac{1}{6} Q^2$. Therefore, total costs equal $\frac{1}{6} Q^2 + (\alpha_2 + \gamma) Q$.

Total revenue is given by selling price (i.e., $\alpha_1 - \frac{1}{3} Q$) times the quantity sold. Therefore, total revenue equals $- \frac{1}{3} Q^2 + \alpha_1 Q$.

Profit $\pi$ is defined as total revenue minus total costs. Thus, $\pi = -\frac{1}{3} Q^2 + \alpha_1 Q - \frac{1}{6} Q^2 - (\alpha_2 + \gamma) = -\frac{1}{2} Q^2 + (\alpha_1 - \alpha_2 - \gamma)Q$.

(b) Clearly, $\pi$ is quadratic function in $Q$ with $a=-\frac{1}{2}$, $b=\alpha_1 - \alpha_2 - \gamma$, and $c=0$. Since $a<0$, the function is a parabola opening downwards. Thus, $Q^* = - \frac{b}{2a} = \alpha_1 - \alpha_2 - \gamma$ is the global maximum point. Thus, if $\alpha_1 - \alpha_2 - \gamma \leq 0$, then $Q^* \leq 0$. This is a nonsense solution: it would suggest buying, shipping, and selling negative quantities. A more reasonable interpretation would be to say that (1) $Q$ cannot be negative and (2) transport and buying costs are so high, that the best choice is to set $Q=0$, in which case $\pi=0$.

(c) Similar to the shipping cost, the total tax is given by $\tau Q$. These costs need to be subtracted from the profit. Thus, $\pi$ after tax is given by $-\frac{1}{2} Q^2 + (\alpha_1 - \alpha_2 - \gamma - \tau) Q$. The optimum for the firm has a similar structure to what we found before, viz., $Q^* = \alpha_1 - \alpha_2 - \gamma - \tau$.

(d) Assuming the firm maximises profit and, thus, always sets $Q = \alpha_1 - \alpha_2 - \gamma - \tau$ tons, the tax revenue ($r$) for the Ghanaian government is given by the tax per ton (i.e., $\tau$) times that amount of tons. Hence, $r = (\alpha_1 - \alpha_2 - \gamma - \tau) \tau = - \tau^2 + (\alpha_1 - \alpha_2 - \gamma) \tau$ is the total export tax revenue for the government.

(e) Clearly, $r$ is a quadratic function with respect to $\tau$, that reaches its maximum at $\tau = \frac{1}{2} \left( \alpha_1 - \alpha_2 - \gamma \right)$.

Question 17

Let $a_1 ,a_2 , \ldots ,a_n$ and $b_1 ,b_2 , \ldots, b_n$ be arbitrary real numbers. The inequality $(a_1 b_1 + a_2 b_2 + \ldots + a_n b_n)^2 ≤ (a_1^2 + a_2^2 + \ldots + a_n^2)(b_1^2 + b_2^2 + \ldots + b_n^2)$ is called the Cauchy–Schwarz inequality.

(a) Check the inequality for $n = 2$, when $a_1 = −3, a_2 = 2, b_1 = 5,$ and $b_2 = −2$.

(b) Prove the inequality by means of the following trick:

• First, define $f$ for all $x$ by $f(x) = (a_1 x + b_1 )^2 + \ldots + (a_n x + b_n )^2$. It should be obvious that $f (x) \geq 0$ for all $x$.

• Write $f (x)$ as $Ax^2 + Bx + C$, where the expressions for the coefficients $A, B,$ and $C$ are related to the terms in the Cauchy-Schwarz inequality. Because $Ax^2 + Bx + C \geq 0$ for all $x$, we must have $B^2 - 4AC \leq 0$. Why?

(c) Show that the Cauchy-Schwarz inequality directly follows from this.

Show answer

(a) In this case, we have $( -3 \cdot 5 + 2 \cdot -2)^2 = (-19)^2 = 361$ on the left-hand side (LHS) and $(-3^2 + 2^2) \cdot (5^2 + (-2)^2) = (9 + 4) \cdot (25 + 4) = 13 \cdot 29 = 377$ on the right-hand side (RHS). Clearly, $361 \leq 377$. So the Cauchy-Schwarz inequality holds true for the particular numbers considered here. Now let's move to the general case.

(b) In the function $f$ outlined in this subquestion, we effectively take the sum of $(a_i x + b_i)^2$ for $i=1,2,\ldots,n$. That is, we can write $f(x) = \sum_{i=1}^n (a_i x + b_i)^2$. Here, notice that $(a_i x + b_i)^2 = a_i^2 x^2 + 2 a_i b_i x + b_i^2$. Thus, we can write $f(x) = \sum_{i=1}^n a_i^2 x^2 + 2 a_i b_i x + b_i^2$.

Using properties of the sum operator, this can be further rearranged as $f(x) = \left(\sum_{i=1}^n a_i^2 \right) x^2 + 2 \left( \sum_{i=1}^n a_i b_i \right) x + \sum_{i=1}^n b_i^2$. Thus, we can write $f(x)$ in the desired form for this subquestion by setting $A = \sum_{i=1}^n a_i^2$, $B=2 \sum_{i=1}^n a_i b_i$, and $C= \sum_{i=1}^n b_i^2$, which reveals that $f(x)$ is a parabola.

Going back to the formulation in which we state that $f(x) = \sum_{i=1}^n (a_i x + b_i)^2$, notice that $(a_i x + b_i)^2 \geq 0$ for any $x \in \mathbb{R}$, because squaring the output of $a_i x + b_i$ always yields a nonnegative number. Hence, each term in the sum is always nonnegative, and, therefore, $f(x)$ itself is always nonnegative.

Since $f(x)$ is a parabola that is nonnegative, it either has exactly one point where it touches the horizontal axis (i.e., $y=0$), in which case the discriminant (here given by $B^2 - 4AC$) is zero, or it has no points at all where it touches the horizontal axis, in which case the discriminant is negative. Thus, positive values of the discriminant cannot arise, because that would yield two roots and values for $x$ where $f < 0$ (which contradicts what we established earlier: $f \geq 0$). Thus, we have shown that $B^2 - 4AC \leq 0$.

(c) Plugging our expressions for $A$, $B$, and $C$ into the inequality for the discriminant, yields the following inequality: $4 \left( \sum_{i=1}^n a_i b_i \right)^2 - 4 \left( \sum_{i=1}^n a_i^2 \right) \left( \sum_{i=1}^n b_i^2 \right) \leq 0$. Dividing both sides of the inequality by $4$ and then adding $\left( \sum_{i=1}^n a_i^2 \right) \left( \sum_{i=1}^n b_i^2 \right)$ to both sides yields $\left( \sum_{i=1}^n a_i b_i \right)^2 \leq \left( \sum_{i=1}^n a_i^2 \right) \left( \sum_{i=1}^n b_i^2 \right)$.

Writing out this inequality element-wise (i.e., without using the summation operator), the inequality states that $(a_1 b_1 + \ldots + a_n b_n)^2 \leq (a_1^2 + \ldots + a_n^2) \cdot (b_1^2 + \ldots + b_n^2 )$, which is the Cauchy-Schwarz inequality!

To recap how you proved this: you were presented with a smartly chosen function $f$. For that function, you were able to prove it was a nonnegative quadratic function. With that insight in mind, you were able to argue that the discriminant $d$ of that function cannot be positive (i.e., $d \leq 0$). By rearranging this inequality, you established the Cauchy-Schwarz inequality.