Answer :
Sure! Let's go through each function step-by-step and find the images of the given values [tex]\([-3, -1, 0, 1, 2, 4]\)[/tex].
### (a) [tex]\( f(x) = x^2 - x - 5 \)[/tex]
To find the images of [tex]\( f(x) \)[/tex], we will substitute the values into the function [tex]\( f(x) \)[/tex].
1. For [tex]\( x = -3 \)[/tex]:
[tex]\[ f(-3) = (-3)^2 - (-3) - 5 = 9 + 3 - 5 = 7 \][/tex]
2. For [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = (-1)^2 - (-1) - 5 = 1 + 1 - 5 = -3 \][/tex]
3. For [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 0^2 - 0 - 5 = -5 \][/tex]
4. For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = 1^2 - 1 - 5 = 1 - 1 - 5 = -5 \][/tex]
5. For [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = 2^2 - 2 - 5 = 4 - 2 - 5 = -3 \][/tex]
6. For [tex]\( x = 4 \)[/tex]:
[tex]\[ f(4) = 4^2 - 4 - 5 = 16 - 4 - 5 = 7 \][/tex]
So, the images of [tex]\( f(x) \)[/tex] for [tex]\([-3, -1, 0, 1, 2, 4]\)[/tex] are:
[tex]\[ \boxed{[7, -3, -5, -5, -3, 7]} \][/tex]
### (b) [tex]\( g(x) = (x+1)^2 \)[/tex]
To find the images of [tex]\( g(x) \)[/tex], we will substitute the values into the function [tex]\( g(x) \)[/tex].
1. For [tex]\( x = -3 \)[/tex]:
[tex]\[ g(-3) = (-3+1)^2 = (-2)^2 = 4 \][/tex]
2. For [tex]\( x = -1 \)[/tex]:
[tex]\[ g(-1) = (-1+1)^2 = 0^2 = 0 \][/tex]
3. For [tex]\( x = 0 \)[/tex]:
[tex]\[ g(0) = (0+1)^2 = 1^2 = 1 \][/tex]
4. For [tex]\( x = 1 \)[/tex]:
[tex]\[ g(1) = (1+1)^2 = 2^2 = 4 \][/tex]
5. For [tex]\( x = 2 \)[/tex]:
[tex]\[ g(2) = (2+1)^2 = 3^2 = 9 \][/tex]
6. For [tex]\( x = 4 \)[/tex]:
[tex]\[ g(4) = (4+1)^2 = 5^2 = 25 \][/tex]
So, the images of [tex]\( g(x) \)[/tex] for [tex]\([-3, -1, 0, 1, 2, 4]\)[/tex] are:
[tex]\[ \boxed{[4, 0, 1, 4, 9, 25]} \][/tex]
### (c) [tex]\( h(x) = x - 1 \)[/tex]
To find the images of [tex]\( h(x) \)[/tex], we will substitute the values into the function [tex]\( h(x) \)[/tex].
1. For [tex]\( x = -3 \)[/tex]:
[tex]\[ h(-3) = -3 - 1 = -4 \][/tex]
2. For [tex]\( x = -1 \)[/tex]:
[tex]\[ h(-1) = -1 - 1 = -2 \][/tex]
3. For [tex]\( x = 0 \)[/tex]:
[tex]\[ h(0) = 0 - 1 = -1 \][/tex]
4. For [tex]\( x = 1 \)[/tex]:
[tex]\[ h(1) = 1 - 1 = 0 \][/tex]
5. For [tex]\( x = 2 \)[/tex]:
[tex]\[ h(2) = 2 - 1 = 1 \][/tex]
6. For [tex]\( x = 4 \)[/tex]:
[tex]\[ h(4) = 4 - 1 = 3 \][/tex]
So, the images of [tex]\( h(x) \)[/tex] for [tex]\([-3, -1, 0, 1, 2, 4]\)[/tex] are:
[tex]\[ \boxed{[-4, -2, -1, 0, 1, 3]} \][/tex]
### (d) [tex]\( F(x) = (x+1)(x-2) \)[/tex]
To find the images of [tex]\( F(x) \)[/tex], we will substitute the values into the function [tex]\( F(x) \)[/tex].
1. For [tex]\( x = -3 \)[/tex]:
[tex]\[ F(-3) = (-3+1)(-3-2) = (-2)(-5) = 10 \][/tex]
2. For [tex]\( x = -1 \)[/tex]:
[tex]\[ F(-1) = (-1+1)(-1-2) = 0 \cdot (-3) = 0 \][/tex]
3. For [tex]\( x = 0 \)[/tex]:
[tex]\[ F(0) = (0+1)(0-2) = 1 \cdot (-2) = -2 \][/tex]
4. For [tex]\( x = 1 \)[/tex]:
[tex]\[ F(1) = (1+1)(1-2) = 2 \cdot (-1) = -2 \][/tex]
5. For [tex]\( x = 2 \)[/tex]:
[tex]\[ F(2) = (2+1)(2-2) = 3 \cdot 0 = 0 \][/tex]
6. For [tex]\( x = 4 \)[/tex]:
[tex]\[ F(4) = (4+1)(4-2) = 5 \cdot 2 = 10 \][/tex]
So, the images of [tex]\( F(x) \)[/tex] for [tex]\([-3, -1, 0, 1, 2, 4]\)[/tex] are:
[tex]\[ \boxed{[10, 0, -2, -2, 0, 10]} \][/tex]
### (a) [tex]\( f(x) = x^2 - x - 5 \)[/tex]
To find the images of [tex]\( f(x) \)[/tex], we will substitute the values into the function [tex]\( f(x) \)[/tex].
1. For [tex]\( x = -3 \)[/tex]:
[tex]\[ f(-3) = (-3)^2 - (-3) - 5 = 9 + 3 - 5 = 7 \][/tex]
2. For [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = (-1)^2 - (-1) - 5 = 1 + 1 - 5 = -3 \][/tex]
3. For [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 0^2 - 0 - 5 = -5 \][/tex]
4. For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = 1^2 - 1 - 5 = 1 - 1 - 5 = -5 \][/tex]
5. For [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = 2^2 - 2 - 5 = 4 - 2 - 5 = -3 \][/tex]
6. For [tex]\( x = 4 \)[/tex]:
[tex]\[ f(4) = 4^2 - 4 - 5 = 16 - 4 - 5 = 7 \][/tex]
So, the images of [tex]\( f(x) \)[/tex] for [tex]\([-3, -1, 0, 1, 2, 4]\)[/tex] are:
[tex]\[ \boxed{[7, -3, -5, -5, -3, 7]} \][/tex]
### (b) [tex]\( g(x) = (x+1)^2 \)[/tex]
To find the images of [tex]\( g(x) \)[/tex], we will substitute the values into the function [tex]\( g(x) \)[/tex].
1. For [tex]\( x = -3 \)[/tex]:
[tex]\[ g(-3) = (-3+1)^2 = (-2)^2 = 4 \][/tex]
2. For [tex]\( x = -1 \)[/tex]:
[tex]\[ g(-1) = (-1+1)^2 = 0^2 = 0 \][/tex]
3. For [tex]\( x = 0 \)[/tex]:
[tex]\[ g(0) = (0+1)^2 = 1^2 = 1 \][/tex]
4. For [tex]\( x = 1 \)[/tex]:
[tex]\[ g(1) = (1+1)^2 = 2^2 = 4 \][/tex]
5. For [tex]\( x = 2 \)[/tex]:
[tex]\[ g(2) = (2+1)^2 = 3^2 = 9 \][/tex]
6. For [tex]\( x = 4 \)[/tex]:
[tex]\[ g(4) = (4+1)^2 = 5^2 = 25 \][/tex]
So, the images of [tex]\( g(x) \)[/tex] for [tex]\([-3, -1, 0, 1, 2, 4]\)[/tex] are:
[tex]\[ \boxed{[4, 0, 1, 4, 9, 25]} \][/tex]
### (c) [tex]\( h(x) = x - 1 \)[/tex]
To find the images of [tex]\( h(x) \)[/tex], we will substitute the values into the function [tex]\( h(x) \)[/tex].
1. For [tex]\( x = -3 \)[/tex]:
[tex]\[ h(-3) = -3 - 1 = -4 \][/tex]
2. For [tex]\( x = -1 \)[/tex]:
[tex]\[ h(-1) = -1 - 1 = -2 \][/tex]
3. For [tex]\( x = 0 \)[/tex]:
[tex]\[ h(0) = 0 - 1 = -1 \][/tex]
4. For [tex]\( x = 1 \)[/tex]:
[tex]\[ h(1) = 1 - 1 = 0 \][/tex]
5. For [tex]\( x = 2 \)[/tex]:
[tex]\[ h(2) = 2 - 1 = 1 \][/tex]
6. For [tex]\( x = 4 \)[/tex]:
[tex]\[ h(4) = 4 - 1 = 3 \][/tex]
So, the images of [tex]\( h(x) \)[/tex] for [tex]\([-3, -1, 0, 1, 2, 4]\)[/tex] are:
[tex]\[ \boxed{[-4, -2, -1, 0, 1, 3]} \][/tex]
### (d) [tex]\( F(x) = (x+1)(x-2) \)[/tex]
To find the images of [tex]\( F(x) \)[/tex], we will substitute the values into the function [tex]\( F(x) \)[/tex].
1. For [tex]\( x = -3 \)[/tex]:
[tex]\[ F(-3) = (-3+1)(-3-2) = (-2)(-5) = 10 \][/tex]
2. For [tex]\( x = -1 \)[/tex]:
[tex]\[ F(-1) = (-1+1)(-1-2) = 0 \cdot (-3) = 0 \][/tex]
3. For [tex]\( x = 0 \)[/tex]:
[tex]\[ F(0) = (0+1)(0-2) = 1 \cdot (-2) = -2 \][/tex]
4. For [tex]\( x = 1 \)[/tex]:
[tex]\[ F(1) = (1+1)(1-2) = 2 \cdot (-1) = -2 \][/tex]
5. For [tex]\( x = 2 \)[/tex]:
[tex]\[ F(2) = (2+1)(2-2) = 3 \cdot 0 = 0 \][/tex]
6. For [tex]\( x = 4 \)[/tex]:
[tex]\[ F(4) = (4+1)(4-2) = 5 \cdot 2 = 10 \][/tex]
So, the images of [tex]\( F(x) \)[/tex] for [tex]\([-3, -1, 0, 1, 2, 4]\)[/tex] are:
[tex]\[ \boxed{[10, 0, -2, -2, 0, 10]} \][/tex]
