# Solved Examples on the Properties of Relations and Functions

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(1.) Write these inequalities in interval notation.

$(a.)\;\;-3 \lt x \lt 5 \\[3ex] (b.)\;\;-3 \le x \lt 5 \\[3ex] (c.)\;\;-3 \lt x \le 5 \\[3ex] (d.)\;\;-3 \le x \le 5 \\[3ex] (e.)\;\; x \lt 5 \\[3ex] (f.)\;\; x \gt 5 \\[3ex] (g.)\;\; x \le 5 \\[3ex] (h.)\;\; 5 \ge x \\[3ex] (i)\;\; 5 \gt x \\[3ex] (j.)\;\; 5 \lt x \\[3ex] (k.)\;\; 5 \le x \\[3ex]$

$(a.) \\[3ex] -3 \lt x \lt 5 \\[3ex] (-3, 5) \\[3ex] (b.) \\[3ex] -3 \le x \lt 5 \\[3ex] [-3, 5) \\[3ex] (c.) \\[3ex] -3 \lt x \le 5 \\[3ex] (-3, 5] \\[3ex] (d.) \\[3ex] -3 \le x \le 5 \\[3ex] [-3, 5] \\[3ex] (e.) \\[3ex] x \lt 5 \\[3ex] (-\infty, 5) \\[3ex] (f.) \\[3ex] x \gt 5 \\[3ex] (5, \infty) \\[3ex] (g.) \\[3ex] x \le 5 \\[3ex] (-\infty, 5] \\[3ex] (h.) \\[3ex] 5 \ge x \;\;means\;\;that\;\; x \le 5 \\[3ex] (-\infty, 5] \\[3ex] (i.) \\[3ex] 5 \gt x \implies x \lt 5 \\[3ex] (-\infty, 5) \\[3ex] (j.) \\[3ex] 5 \lt x \implies x \gt 5 \\[3ex] (5, \infty) \\[3ex] (k.) \\[3ex] 5 \le x \implies x \ge 5 \\[3ex] [5, \infty)$
(2.) Write these intervals in set notation:

$(a.)\;\; (7, 9) \\[3ex]$

$(a.) \\[3ex] (7,9) = \{x| 7 \lt x \lt 9\} \\[3ex]$
(3.) (a.) Plot the points (3, −1), (6, −9), (9, −6), (3, −5) in the Cartesian plane.
Choose the correct graph.

(b.) Does the relation represent a​ function? Why?

A. Yes, because there are ordered pairs with the same first element and different second elements.
B. Yes, because there are no ordered pairs with the same first element and different second elements.
C. No, because there are no ordered pairs with the same first element and different second elements.
D. No, because there are ordered pairs with the same first element and different second elements.

(a.) Scale: 1 cm represents 2 units (Not drawn to scale)
The correct graph representing the points is Option D.:

(b.) The relation is not a function because there is an input, 3 that has two different outputs, −1 and −5
D. No, because there are ordered pairs with the same first element and different second elements.
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(6.) Which of the following intervals is required to guarantee a continuous function will have both an absolute maximum and an absolute​ minimum?

$A.\;\; (a, b) \\[3ex] B.\;\; [a, b] \\[3ex] C.\;\; [a, b) \\[3ex] D.\;\; (a, b] \\[3ex]$

The correct option is B.
If f is a continuous​ function, whose domain is a closed interval​ [a,b] , then f has an absolute maximum and an absolute minimum on​ [a,b].
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(9.) Determine the intercepts of these equations:

$(a.)\;\; y = 6x^2 - 6 \\[3ex] (b.)\;\; 9x^2 + 4y^2 = 36 \\[3ex]$

To find: (1.) the x-intercept, set y = 0 and solve for x
(x-values, 0) gives the x-intercept

(2.) the y-intercept, set x = 0 and solve for y
(0, y-values) gives the y-intercept

$(a.) \\[3ex] y = 8x^2 - 8 \\[5ex] \underline{y-intercept} \\[3ex] y = 8(0)^2 - 8 \\[3ex] y = 0 - 8 \\[3ex] y = -8 \\[3ex] y-intercept = (0, -8) \\[3ex] \underline{x-intercept} \\[3ex] 0 = 8x^2 - 8 \\[3ex] 8x^2 - 8 = 0 \\[3ex] 8(x^2 - 1) = 0 \\[3ex] x^2 - 1 = 0 \\[3ex] x^2 = 0 + 1 \\[3ex] x^2 = 1 \\[3ex] x = \pm\sqrt{1} \\[3ex] x = \pm 1 \\[3ex] x-intercepts = (-1, 0) \;\;\;and\;\;\; (1, 0) \\[5ex] (b.) \\[3ex] 9x^2 + 4y^2 = 36 \\[3ex] \underline{y-intercept} \\[3ex] 9(0)^2 + 4y^2 = 36 \\[3ex] 0 + 4y^2 = 36 \\[3ex] 4y^2 = 36 \\[3ex] y^2 = \dfrac{36}{4} \\[5ex] y^2 = 9 \\[3ex] y = \pm \sqrt{9} \\[3ex] y = \pm 3 \\[3ex] y-intercepts = (0, -3) \;\;\;and\;\;\; (0, 3) \\[3ex] \underline{x-intercept} \\[3ex] 9x^2 + 4(0)^2 = 36 \\[3ex] 9x^2 + 0 = 36 \\[3ex] 9x^2 = 36 \\[3ex] x^2 = \dfrac{36}{9} \\[5ex] x^2 = 4 \\[3ex] x = \pm \sqrt{4} \\[3ex] x-intercepts = (-2, 0) \;\;\;and\;\;\; (2, 0) \\[5ex]$
(10.)

(11.) Determine whether the graph is that of a function by using the​ vertical-line test.
If it is a function, use the graph to find
​(a.) its domain and range.
​(b.) the​ intercepts, if any.
​(c.) any symmetry with respect to the​ x-axis, y-axis, or the origin.

The graph failed the Vertical Line Test because the vertical line: $x = 6$ intersects the graph at two points $(6, 2)$ and $(6, -2)$
The same input, 6 has two different outputs: 2 and −2
It is not a function.
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(15.) Determine whether the graph is that of a function by using the​ vertical-line test.
If it is a function, use the graph to find
​(a.) the domain and range.
​(b.) the​ intercepts, if any.
​(c.) any symmetry with respect to the​ x-axis, y-axis, or the origin.

The graph passed the Vertical Line Test because the vertical line: $x = \dfrac{\pi}{2}$ intersects the graph at only one point $\left(\dfrac{\pi}{2}, -1\right)$

It is a function.
It is a cubic function.
It is a polynomial function.

$(a.) \\[3ex] \underline{Domain} \\[3ex] Minimum\;\;x-value = -\pi \\[3ex] Maximum x-value = \pi \\[3ex] D = [-\pi, \pi] \\[3ex] \underline{Range} \\[3ex] Minimum\;\;y-value = -1 \\[3ex] Maximum\;\;y-value = 1 \\[3ex] R = [-1, 1] \\[3ex] (b.) \\[3ex] \underline{x-intercepts} \\[3ex] x-intercepts = (-\pi, 0) \;\;\;and\;\;\; (\pi, 0) \\[3ex] \underline{y-intercept} \\[3ex] y-intercept = (0, 0) \\[3ex]$ (c.) Let us check each symmetry.
For each $(x, y)$ on the graph; there is also $(x, -y)$ on the same graph
Same $x$, Opposite $y$
We have $\left(\dfrac{\pi}{2}, -1\right)$ but we do not have $\left(\dfrac{\pi}{2}, 1\right)$ on the graph.
The graph is not symmetric about the x-axis

For each $(x, y)$ on the graph; there is also $(−x, y)$ on the same graph
Same $y$, Opposite $x$
We have $\left(\dfrac{\pi}{2}, -1\right)$ but we do not have $\left(-\dfrac{\pi}{2}, -1\right)$ on the graph.
The graph is not symmetric about the y-axis

For each $(x, y)$ on the graph; there is also $(−x, −y)$ on the same graph
Opposite $x$, Opposite $y$
We have $\left(\dfrac{\pi}{2}, -1\right)$ and we also have $\left(-\dfrac{\pi}{2}, 1\right)$ on the graph.
The graph is symmetric about the origin.
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(23.) Determine whether the graph is that of a function by using the​ vertical-line test.
If it is a function, use the graph to find
​(a.) its domain and range.
​(b.) the​ intercepts, if any.
​(c.) any symmetry with respect to the​ x-axis, y-axis, or the origin.

The graph failed the Vertical Line Test because the vertical line: $x = -1$ intersects the graph at two points $(-1, 1)$ and $(-1, -1)$
The same input, −1 has two different outputs: 1 and −1
It is not a function.
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(27.) Determine whether the graph is that of a function by using the​ vertical-line test.
If it is a function, use the graph to find
​(a.) the domain and range (Assume that the curve approaches but never intersects the​ y-axis).
​(b.) the​ intercepts, if any.
​(c.) any symmetry with respect to the​ x-axis, y-axis, or the origin.

The graph passed the Vertical Line Test because the vertical line: $x = 1$ intersects the graph at only one point $(1, 0)$
It is a function.

$(a.) \\[3ex] \underline{Domain} \\[3ex] Minimum\;\;x-value = 0(excluded) \\[3ex] Maximum x-value = 2(excluded) \\[3ex] D = (0, 2) \\[3ex] \underline{Range} \\[3ex] Minimum\;\;y-value = -\infty(excluded) \\[3ex] Maximum\;\;y-value = 2 \\[3ex] R = (-\infty, 2) \\[3ex] (b.) \\[3ex] \underline{x-intercept} \\[3ex] x-intercept = (1, 0) \\[3ex] \underline{y-intercept} \\[3ex] y-intercept = None \\[3ex]$ (c.) Let us check each symmetry.
For each $(x, y)$ on the graph; there is also $(x, -y)$ on the same graph
Same $x$, Opposite $y$
We have $(2, 2)$ but we do not have $(2, -2)$ on the graph.
The graph is not symmetric about the x-axis

For each $(x, y)$ on the graph; there is also $(−x, y)$ on the same graph
Same $y$, Opposite $x$
We have $(1, 0)$ but we do not have $(-1, 0)$ on the graph.
The graph is not symmetric about the y-axis

For each $(x, y)$ on the graph; there is also $(−x, −y)$ on the same graph
Opposite $x$, Opposite $y$
We have $(2, 2)$ but we do not have $(-2, -2)$ on the graph.
The graph is not symmetric about the origin.

Therefore, the graph has no symmetry.
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