The table of values for the functions f(x) = x^2 + 1, g(x) = f(x - 5) and h(x) = f(x+3) is:
[tex]\left[\begin{array}{cccccc}x&f(x)&x&g(x)&x&h(x)\\-2&5&-2&50&-2&2\\-1&2&-1&37&-1&5&0&1&0&26&0&10&1&2&1&17&1&17&2&5&2&10&2&26\end{array}\right][/tex]
How to set up the table
The equations of the functions are given as:
[tex]f(x) = x^2 + 1[/tex]
[tex]g(x) = f(x - 5)[/tex]
[tex]h(x) = f(x+3)[/tex]
For function f(x), we have:
f(-2) = (-2)^2 + 1 = 5
f(-1) = (-1)^2 + 1 = 2
f(0) = (0)^2 + 1 = 1
f(1) = (1)^2 + 1 = 2
f(2) = (2)^2 + 1 = 5
For function g(x), we have:
g(-2) = f(-2 - 5) = f(-7) = (-7)^2 + 1 = 50
g(-1) = f(-1 - 5) = f(-6) = (-6)^2 + 1 = 37
g(0) = f(0 - 5) = f(-5) = (-5)^2 + 1 = 26
g(1) = f(1 - 5) = f(-4) = (-4)^2 + 1 = 17
g(2) = f(2 - 5) = f(-3) = (-3)^2 + 1 = 10
For function h(x), we have:
h(-2) = f(-2 + 3) = f(1) = (1)^2 + 1 = 2
h(-1) = f(-1 + 3) = f(2) = (2)^2 + 1 = 5
h(0) = f(0 + 3) = f(3) = (3)^2 + 1 = 10
h(1) = f(1 + 3) = f(4) = (4)^2 + 1 = 17
h(2) = f(2 + 3) = f(5) = (5)^2 + 1 = 26
Substitute the above values in the blank table.
So, we have:
[tex]\left[\begin{array}{cccccc}x&f(x)&x&g(x)&x&h(x)\\-2&5&-2&50&-2&2\\-1&2&-1&37&-1&5&0&1&0&26&0&10&1&2&1&17&1&17&2&5&2&10&2&26\end{array}\right][/tex]
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