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1. For each of the following functions, find the images of [tex]\(-3, -1, 0, 1, 2, 4\)[/tex]:

(a) [tex]\(f(x) = x^2 - x - 5\)[/tex]

(b) [tex]\(g(x) = (x + 1)^2\)[/tex]

(c) [tex]\(h(x) = \frac{x - 1}{2}\)[/tex]

(d) [tex]\(F(x) = (x + 1)(x - 2)\)[/tex]

2. What value of [tex]\(x\)[/tex] must be excluded from the domain of the function in Question 1 part (c)?



Answer :

GinnyAnswer
Let's proceed step-by-step to solve each part of the question.

### 1. Finding the images for the given functions:

a. For the function [tex]\( f(x) = x^2 - x - 5 \)[/tex]:
- When [tex]\( x = -3 \)[/tex]:
[tex]\[ f(-3) = (-3)^2 - (-3) - 5 = 9 + 3 - 5 = 7 \][/tex]
- When [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = (-1)^2 - (-1) - 5 = 1 + 1 - 5 = -3 \][/tex]
- When [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 0^2 - 0 - 5 = -5 \][/tex]
- When [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = 1^2 - 1 - 5 = 1 - 1 - 5 = -5 \][/tex]
- When [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = 2^2 - 2 - 5 = 4 - 2 - 5 = -3 \][/tex]
- When [tex]\( x = 4 \)[/tex]:
[tex]\[ f(4) = 4^2 - 4 - 5 = 16 - 4 - 5 = 7 \][/tex]

The images of [tex]\(-3, -1, 0, 1, 2, 4\)[/tex] under [tex]\( f(x) \)[/tex] are: [tex]\([7, -3, -5, -5, -3, 7]\)[/tex].

b. For the function [tex]\( g(x) = (x + 1)^2 \)[/tex]:
- When [tex]\( x = -3 \)[/tex]:
[tex]\[ g(-3) = (-3 + 1)^2 = (-2)^2 = 4 \][/tex]
- When [tex]\( x = -1 \)[/tex]:
[tex]\[ g(-1) = (-1 + 1)^2 = 0^2 = 0 \][/tex]
- When [tex]\( x = 0 \)[/tex]:
[tex]\[ g(0) = (0 + 1)^2 = 1^2 = 1 \][/tex]
- When [tex]\( x = 1 \)[/tex]:
[tex]\[ g(1) = (1 + 1)^2 = 2^2 = 4 \][/tex]
- When [tex]\( x = 2 \)[/tex]:
[tex]\[ g(2) = (2 + 1)^2 = 3^2 = 9 \][/tex]
- When [tex]\( x = 4 \)[/tex]:
[tex]\[ g(4) = (4 + 1)^2 = 5^2 = 25 \][/tex]

The images of [tex]\(-3, -1, 0, 1, 2, 4\)[/tex] under [tex]\( g(x) \)[/tex] are: [tex]\([4, 0, 1, 4, 9, 25]\)[/tex].

c. For the function [tex]\( h(x) = \frac{x-1}{2} \)[/tex]:
- When [tex]\( x = -3 \)[/tex]:
[tex]\[ h(-3) = \frac{-3 - 1}{2} = \frac{-4}{2} = -2.0 \][/tex]
- When [tex]\( x = -1 \)[/tex]:
[tex]\[ h(-1) = \frac{-1 - 1}{2} = \frac{-2}{2} = -1.0 \][/tex]
- When [tex]\( x = 0 \)[/tex]:
[tex]\[ h(0) = \frac{0 - 1}{2} = \frac{-1}{2} = -0.5 \][/tex]
- When [tex]\( x = 1 \)[/tex]:
[tex]\[ h(1) = \frac{1 - 1}{2} = \frac{0}{2} = 0.0 \][/tex]
- When [tex]\( x = 2 \)[/tex]:
[tex]\[ h(2) = \frac{2 - 1}{2} = \frac{1}{2} = 0.5 \][/tex]
- When [tex]\( x = 4 \)[/tex]:
[tex]\[ h(4) = \frac{4 - 1}{2} = \frac{3}{2} = 1.5 \][/tex]

The images of [tex]\(-3, -1, 0, 1, 2, 4\)[/tex] under [tex]\( h(x) \)[/tex] are: [tex]\([-2.0, -1.0, -0.5, 0.0, 0.5, 1.5]\)[/tex].

d. For the function [tex]\( F(x) = (x + 1)(x - 2) \)[/tex]:
- When [tex]\( x = -3 \)[/tex]:
[tex]\[ F(-3) = (-3 + 1)(-3 - 2) = (-2)(-5) = 10 \][/tex]
- When [tex]\( x = -1 \)[/tex]:
[tex]\[ F(-1) = (-1 + 1)(-1 - 2) = (0)(-3) = 0 \][/tex]
- When [tex]\( x = 0 \)[/tex]:
[tex]\[ F(0) = (0 + 1)(0 - 2) = (1)(-2) = -2 \][/tex]
- When [tex]\( x = 1 \)[/tex]:
[tex]\[ F(1) = (1 + 1)(1 - 2) = (2)(-1) = -2 \][/tex]
- When [tex]\( x = 2 \)[/tex]:
[tex]\[ F(2) = (2 + 1)(2 - 2) = (3)(0) = 0 \][/tex]
- When [tex]\( x = 4 \)[/tex]:
[tex]\[ F(4) = (4 + 1)(4 - 2) = (5)(2) = 10 \][/tex]

The images of [tex]\(-3, -1, 0, 1, 2, 4\)[/tex] under [tex]\( F(x) \)[/tex] are: [tex]\([10, 0, -2, -2, 0, 10]\)[/tex].

### 2. Excluded Values for [tex]\( h(x) = \frac{x-1}{2} \)[/tex]:

The function [tex]\( h(x) \)[/tex] is undefined when the denominator is zero.
[tex]\((x-1)/2\)[/tex] has a denominator of 2, which is never zero. Thus, there are no [tex]\( x \)[/tex]-values that need to be excluded from the domain.

The value of [tex]\( x \)[/tex] that must be excluded from the domain of [tex]\( h(x) \)[/tex] is: [tex]\(\text{None}\)[/tex].

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