Sure, let us solve this step-by-step.
Given Function:
[tex]\[ f(x) = 3x^2 + 2 \][/tex]
Known Values:
[tex]\[ a = 2 \][/tex]
[tex]\[ h = 1 \][/tex]
### Part (a) - Find [tex]\( f(a) \)[/tex]
To find [tex]\( f(a) \)[/tex], we substitute [tex]\( a = 2 \)[/tex] into the function [tex]\( f(x) \)[/tex]:
[tex]\[ f(a) = f(2) = 3(2)^2 + 2 \][/tex]
[tex]\[ f(2) = 3 \cdot 4 + 2 \][/tex]
[tex]\[ f(2) = 12 + 2 \][/tex]
[tex]\[ f(2) = 14 \][/tex]
So, [tex]\( f(a) = 14 \)[/tex].
### Part (b) - Find [tex]\( f(a+h) \)[/tex]
To find [tex]\( f(a+h) \)[/tex], we substitute [tex]\( a = 2 \)[/tex] and [tex]\( h = 1 \)[/tex] into the function [tex]\( f(x) \)[/tex]:
[tex]\[ f(a + h) = f(2 + 1) = f(3) \][/tex]
[tex]\[ f(3) = 3(3)^2 + 2 \][/tex]
[tex]\[ f(3) = 3 \cdot 9 + 2 \][/tex]
[tex]\[ f(3) = 27 + 2 \][/tex]
[tex]\[ f(3) = 29 \][/tex]
So, [tex]\( f(a+h) = 29 \)[/tex].
### Part (c) - Find [tex]\( \frac{f(a+h) - f(a)}{h} \)[/tex]
To find the difference quotient, we use the values for [tex]\( f(a) \)[/tex] and [tex]\( f(a+h) \)[/tex]:
[tex]\[ \frac{f(a+h) - f(a)}{h} = \frac{f(3) - f(2)}{1} \][/tex]
[tex]\[ \frac{f(3) - f(2)}{1} = \frac{29 - 14}{1} \][/tex]
[tex]\[ \frac{29 - 14}{1} = \frac{15}{1} \][/tex]
[tex]\[ \frac{15}{1} = 15 \][/tex]
So, [tex]\( \frac{f(a+h) - f(a)}{h} = 15 \)[/tex].
### Summary
The answers to the questions are:
a.) [tex]\( f(a) = 14 \)[/tex]
b.) [tex]\( f(a+h) = 29 \)[/tex]
c.) [tex]\( \frac{f(a+h) - f(a)}{h} = 15 \)[/tex]
