Question

F(x) = 4x + 4 g(x) = 3x +7

Answer 1

Answer:

Part 1) $(f+g)(x)=7x+11$

Part 2) $(f-g)(x)=x-3$

Part 3) $(f/g)(x)=\frac{(4x+4)}{(3x+7)}$

Part 4) In interval notation the domain is  (-∞,-7/3) ∪ (-7/3,∞)

Part 5) $(f o g) (x)=12x+32$

Step-by-step explanation:

we have

$f(x)=4x+4$

$g(x)=3x+7$

Part 1) Find (f+g)(x)

we know that

$(f+g)(x)=f(x)+g(x)$

substitute the given functions

$(f+g)(x)=(4x+4)+(3x+7)$

Combine like terms

$(f+g)(x)=7x+11$

Part 2) Find (f-g)(x)

we know that

$(f-g)(x)=f(x)-g(x)$

substitute the given functions

$(f-g)(x)=(4x+4)-(3x+7)$

$(f-g)(x)=4x+4-3x-7$

Combine like terms

$(f-g)(x)=x-3$

Part 3) Find (f/g)(x)

we know that

$(f/g)(x)=\frac{f(x)}{g(x)}$

substitute the given functions

$(f/g)(x)=\frac{(4x+4)}{(3x+7)}$

Part 4) What the domain for the answer in question 3

we have

$(f/g)(x)=\frac{(4x+4)}{(3x+7)}$

we know that

The denominator of the quotient cannot be equal to zero

so

$3x+7=0$

$3x=-7$

$x=-\frac{7}{3}$

The domain is all real numbers except the number x=-7/3

In interval notation the domain is  (-∞,-7/3) ∪ (-7/3,∞)

Part 5) Find (f o g) (x)

we know that

$(f o g) (x)=f(g(x))$

substitute

$f(g(x))=4(3x+7)+4$

$f(g(x))=12x+28+4$

$f(g(x))=12x+32$

therefore

$(f o g) (x)=12x+32$