Solved Examples on the Properties of Relations and Functions

$ (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) $

$ (a.) \\[3ex] (7,9) = \{x| 7 \lt x \lt 9\} \\[3ex] $

(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.

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].

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] $

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.

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.
Symmetric about the x-axis
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

Symmetric about the y-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

Symmetric about the origin
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.

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.

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.
Symmetric about the x-axis
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

Symmetric about the y-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

Symmetric about the origin
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.