Certainly! Let's consider the function [tex]\( g(x) = 3x^2 - 3x + 16 \)[/tex].
### a) [tex]\( g(0) \)[/tex]
To find [tex]\( g(0) \)[/tex], we substitute [tex]\( x = 0 \)[/tex] into the function:
[tex]\[
g(0) = 3(0)^2 - 3(0) + 16 = 16
\][/tex]
So, [tex]\( g(0) = 16 \)[/tex].
### b) [tex]\( g(-1) \)[/tex]
To find [tex]\( g(-1) \)[/tex], we substitute [tex]\( x = -1 \)[/tex] into the function:
[tex]\[
g(-1) = 3(-1)^2 - 3(-1) + 16 = 3(1) + 3 + 16 = 3 + 3 + 16 = 22
\][/tex]
So, [tex]\( g(-1) = 22 \)[/tex].
### c) [tex]\( g(2) \)[/tex]
To find [tex]\( g(2) \)[/tex], we substitute [tex]\( x = 2 \)[/tex] into the function:
[tex]\[
g(2) = 3(2)^2 - 3(2) + 16 = 3(4) - 6 + 16 = 12 - 6 + 16 = 22
\][/tex]
So, [tex]\( g(2) = 22 \)[/tex].
### d) [tex]\( g(-x) \)[/tex]
To find [tex]\( g(-x) \)[/tex], we substitute [tex]\( x = -x \)[/tex] into the function:
[tex]\[
g(-x) = 3(-x)^2 - 3(-x) + 16 = 3x^2 + 3x + 16
\][/tex]
So, [tex]\( g(-x) = 3x^2 + 3x + 16 \)[/tex].
### e) [tex]\( g(1-t) \)[/tex]
To find [tex]\( g(1-t) \)[/tex], we substitute [tex]\( x = 1-t \)[/tex] into the function:
[tex]\[
g(1-t) = 3(1-t)^2 - 3(1-t) + 16
\][/tex]
We first expand [tex]\( (1-t)^2 \)[/tex]:
[tex]\[
(1-t)^2 = 1 - 2t + t^2
\][/tex]
Then we substitute:
[tex]\[
g(1-t) = 3(1 - 2t + t^2) - 3(1-t) + 16 = 3(1) - 3(2t) + 3(t^2) - 3(1) + 3(t) + 16
\][/tex]
[tex]\[
= 3 - 6t + 3t^2 - 3 + 3t + 16 = 3t^2 - 3t + 16
\][/tex]
So, [tex]\( g(1-t) = 3t^2 - 3t + 16 \)[/tex].
### Summary:
- [tex]\( g(0) = 16 \)[/tex]
- [tex]\( g(-1) = 22 \)[/tex]
- [tex]\( g(2) = 22 \)[/tex]
- [tex]\( g(-x) = 3x^2 + 3x + 16 \)[/tex]
- [tex]\( g(1-t) = 3t^2 - 3t + 16 \)[/tex]
