Question

function operations
1、let ( f(x) = 3x^2 - 2x + 6 ) and ( g(x) = 7x - 4 ). identify the rule for ( f(x) + g(x) ). show work.
a ( 21x^3 - 26x^2 - 50x - 24 )
b ( 21x^2 - 21x + 50 )
c ( 3x^2 + 5x + 2 )
d ( 3x^2 + 9x - 2 )

2、part a
identify the rule for ( \frac{f}{g} ) when ( f(x) = -3x - 6 ) and ( g(x) = x^2 - x - 6 ). show work!
a ( \frac{f(x)}{g(x)} = \frac{-2}{(x + 2)} )
c ( \frac{f(x)}{g(x)} = \frac{3}{(x + 2)} )
b ( \frac{f(x)}{g(x)} = \frac{-3(x - 2)}{(x - 6)(x + 1)} )
d ( \frac{f(x)}{g(x)} = \frac{-3}{x - 3} )

part b
select all values that are not in the domain of ( \frac{f}{g} ).
a. ( -1 )
d. ( 6 )
b. ( 3 )
e. ( -2 )
c. ( -3 )

3、let ( h(x) = sqrt{x + 3} ). find the simplified value of ( h(22) ).
a ( 5 )
c ( sqrt{22} )
b ( sqrt{3} )
d ( sqrt{25} )

4、let ( f(x) = 3x^2 + 2 ) and ( g(x) = sqrt{x - 4} ). what is the rule for the composition ( f circ g )?
show work!
a ( (f circ g)(x) = sqrt{3x^2 - 2} ), domain is all real numbers
b ( (f circ g)(x) = sqrt{3x^2 - 2} ), domain is ( x geq 4 )
c ( (f circ g)(x) = 3x - 10 ), domain is all real numbers
d ( (f circ g)(x) = 3x - 10 ), domain is ( x geq 4 )

Explanation:

Question 1

Step1: Add the two functions

To find \( f(x) + g(x) \), we substitute the given functions \( f(x) = 3x^2 - 2x + 6 \) and \( g(x) = 7x - 4 \) and combine like terms.
\[

$$\begin{align*} f(x) + g(x) &= (3x^2 - 2x + 6) + (7x - 4)\\ &= 3x^2 + (-2x + 7x) + (6 - 4)\\ &= 3x^2 + 5x + 2 \end{align*}$$

\]

Step1: Factor both functions

First, factor \( f(x) = -3x - 6 = -3(x + 2) \) and \( g(x) = x^2 - x - 6 \). To factor \( g(x) \), we find two numbers that multiply to \( -6 \) and add to \( -1 \), which are \( -3 \) and \( 2 \). So, \( g(x) = (x - 3)(x + 2) \).

Step2: Divide the factored functions

Now, find \( \frac{f(x)}{g(x)} \):
\[
\frac{f(x)}{g(x)} = \frac{-3(x + 2)}{(x - 3)(x + 2)}
\]
Cancel out the common factor \( (x + 2) \) (assuming \( x
eq -2 \)):
\[
\frac{f(x)}{g(x)} = \frac{-3}{x - 3}
\]
Wait, no, wait. Wait, \( f(x) = -3x -6 = -3(x + 2) \), \( g(x) = (x - 3)(x + 2) \). So when we divide, \( (x + 2) \) cancels, leaving \( \frac{-3}{x - 3} \)? Wait, but let's check the options. Wait, maybe I made a mistake. Wait, \( f(x) = -3x -6 = -3(x + 2) \), \( g(x) = x^2 -x -6 = (x - 3)(x + 2) \). So \( \frac{f(x)}{g(x)} = \frac{-3(x + 2)}{(x - 3)(x + 2)} = \frac{-3}{x - 3} \) (for \( x
eq -2 \) and \( x
eq 3 \)). But looking at the options, option D is \( \frac{f(x)}{g(x)} = \frac{-3}{x - 3} \). Wait, but let's check again. Wait, maybe I factored \( f(x) \) wrong? No, \( -3x -6 = -3(x + 2) \). \( g(x) = (x - 3)(x + 2) \). So dividing, \( (x + 2) \) cancels, so \( \frac{-3}{x - 3} \), which is option D. Wait, but let's check the options again. Option D is \( \frac{f(x)}{g(x)} = \frac{-3}{x - 3} \). So that's correct.
Wait, but let's check the options again. Wait, option A is \( \frac{-2}{x + 2} \), option B is \( \frac{-3(x - 2)}{(x - 6)(x + 1)} \), option C is \( \frac{3}{x + 2} \), option D is \( \frac{-3}{x - 3} \). So yes, D is correct.
Wait, but let's re-express \( f(x) \): \( f(x) = -3x -6 = -3(x + 2) \), \( g(x) = (x - 3)(x + 2) \). So \( \frac{-3(x + 2)}{(x - 3)(x + 2)} = \frac{-3}{x - 3} \), so D is correct.

The domain of \( \frac{f}{g} \) is all real numbers except where \( g(x) = 0 \) (since division by zero is undefined) and where the canceled factor was zero (from the simplification). First, find where \( g(x) = 0 \): \( g(x) = (x - 3)(x + 2) = 0 \), so \( x = 3 \) or \( x = -2 \). Also, when we simplified \( \frac{f(x)}{g(x)} \), we canceled \( (x + 2) \), so \( x
eq -2 \) (even though \( g(x) = 0 \) at \( x = -2 \), we already excluded that). Wait, no, the domain of \( \frac{f}{g} \) is all real numbers except where \( g(x) = 0 \), because \( f(x) \) is defined for all real numbers. So \( g(x) = 0 \) when \( (x - 3)(x + 2) = 0 \), so \( x = 3 \) or \( x = -2 \). Wait, but let's check the options. The options are A. -1, B. 3, C. -3, D. 6, E. -2. So the values not in the domain are \( x = 3 \) (B) and \( x = -2 \) (E). Wait, but let's confirm. The domain of \( \frac{f}{g} \) is \( x
eq 3 \) and \( x
eq -2 \), because \( g(3) = 0 \) and \( g(-2) = 0 \). So the values not in the domain are 3 (B) and -2 (E).

Answer:

C. \( 3x^2 + 5x + 2 \)

Question 2 - Part A