Exercises 2

This page allows you to practice some exercises on Properties of Functions. 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 5.4 - Graphs of Equations - Section 5.5 - Distance in the Plane - Section 5.6 - General Functions

Question 9

Determine the distances between the following pairs of points. (a) $(x, y)$ and $(2x, y + 3)$

(b) $(a,b)$ and $(-a, b)$

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(a) $\sqrt{(2x - x)^{2} + (y + 3 -y)^{2}} = \sqrt{x^{2} + 9}$.

(b) $\sqrt{ (-a - a)^{2} + (b - b)^{2}} = \sqrt{(-2a)^{2}} = \sqrt{4a^2} = 2|a|$.

Question 10

The distance between $(2,4)$ and $(5,y)$ is $\sqrt{13}$. Find $y$, and explain geometrically why there must be two values of $y$.

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We have $(5 - 2)^{2} + (y -4)^{2} = 13$. This can be written as $y^2 - 8y + 16 = 13 - 9$. Collecting terms gives $y^2 - 8y + 12 = 0$, with solutions $y = 2$ and $y = 6$. A geometric explanation is that the circle with centre at $(2,4)$ and radius $\sqrt{13}$ intersects the line $x=5$ at two points.

Question 11

Find the equations of:

(a) The circle with centre at $(2, 3)$ and radius $4$.

(b) The circle with centre at $(2, 5)$ and one point at $(−1, 3)$.

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(a) Since the circle has centre at $(2,3)$, its equation is $(x-2) + (y -3) = 16$, where $16 = 4^{2} = r^{2}$.

(b) Since the circle has centre at $(2,5)$, its equation is $(x-2)^{2} + (y - 5)^{2} = r^{2}$, with $r^2$ being the squared radius. This value can be computed as follows. Since $(-1,3)$ lies on the circle, $(-1-2)^{2} + (3 -5)^{2} = 13 = r^{2}$.

Question 12

To show that the graph of $x^2 + y^2 − 10x + 14y + 58 = 0$ is a circle, we can argue like this:

• First rearrange the equation to read $(x^2 − 10x) + (y^2 + 14y) = -58$.

• Completing the two squares gives: $(x^2 - 10x + 5^2 ) + (y^2 + 14y + 7^2 ) = -58 + 5^2 + 7^2 = 16$.

• Thus the equation becomes $(x - 5)^2 + (y + 7)^2 = 16$, whose graph is a circle with centre $(5, -7)$ and radius $\sqrt{16} = 4$.

Use this method to find the centre and the radius of the two circles with equations:

(a) $x^2 + y^2 + 10x - 6y + 30 = 0$

(b) $3x^2 + 3y^2 + 18x - 24y = -39$

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(a) Completing the square yields $(x+5)^{2} + (y - 3)^{2} = 4$, so the circle has centre at $(-5,3)$ with radius $2$.

(b) $(x+3)^{2} + (y -4)^{2} = 12$, which has centre at $(-3,4)$ and radius $\sqrt{12} = 2\sqrt{3}$.

Question 13

Prove that if the distance from a point $(x,y)$ to the point $(-2,0)$ is twice the distance from $(x,y)$ to $(4,0)$, then $(x,y)$ must lie on the circle with centre $(6,0)$ and radius $4$.

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The condition imposed in the exercise can be expressed as

$\sqrt{(x+2)^{2} + y^{2}} = 2 \cdot \sqrt{(x-4)^{2} + y^{2}}$.

This leads to $x^2 + 4x + 4 + y^{2} = 4 (x^{2} - 8x + 16 + y^2)$ by squaring and writing out the squares under the roots. Now, collecting terms yields

$-3x^{2} + 36x -60 - 3y^2 = 0 \iff x^{2} -12x + 20 + y^2 = 0$.

We can write this in the following way:

$x^{2} - 12x + 36 -16 + y^{2} = 0 \iff (x-6)^{2} + y^{2} = 16 = 4^{2} = r^{2}$,

which is the desired result.

Question 14

Which of the following rules define functions? For those that are functions, determine whether they are one-to-one, and whether they have an inverse. Determine each inverse when it exists.

(a) The rule that assigns to each person in a classroom his or her height.

(b) The rule that assigns to each mother her youngest surviving child.

(c) The rule that assigns the perimeter of a rectangle to its area.

(d) The rule that assigns the surface area of a spherical ball to its volume.

(e) The rule that assigns the pair of numbers $(x + 3, y)$ to the pair of numbers $(x, y)$.

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Actually, every rule defines a function, except for (c). This is because rectangles with equal areas can have different perimeters. For instance, a $9 \times 1$ and $3 \times 3$ rectangle have the same area ($9$), but different perimeters ($20$ and $12$, respectively) which can be calculated as $2(l + w)$, where $l$ denotes length and $w$ width.

• The function in (b) is one-to-one and has an inverse: the rule mapping each youngest child alive today to his/her mother.

• The function in (d) is one-to-one and has an inverse: the rule mapping the surface area to the volume.

• The function in (e) is one-to-one and has an inverse: the rule that maps $(u, v)$ to $(u-3, v)$.

• The function in (a) is many-to-one, in general, and so has no inverse. This is because there might be several people in a classroom with the same height. Thus, a specific height cannot be uniquely mapped to a single person.