Suppose that the functions f and g are defined for all real numbers x as follows:

f (x) = x + 1

g (x) = 4x + 2

Write the expressions for (f times g) (x) and (f-g) (x) and evaluate (f+g) (3).

a) 4x^2 + 6x + 2

b) - 3x - 1

c) 18

Step-by-step explanation:

f (x) = x + 1

g (x) = 4x + 2

a) (f*g) (x)

= f (x) * g (x)

= (x + 1) (4x + 2)

= 4x^2 + 2x + 4x + 2

= 4x^2 + 6x + 2

b) (f-g) (x)

= f (x) - g (x)

= (x + 1) - (4x + 2)

= x + 1 - 4x - 2

= x - 4x + 1 - 2

= - 3x - 1

c) (f+g) (3)

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

= (x + 1) + (4x + 2)

= x + 1 + 4x + 2

= x + 4x + 1 + 2

= 5x + 3

put x = 3

= 5 (3) + 3

= 15 + 3

= 18

1) (fg) (x) = 4x^2 + 6x + 2

2) (f-g) (x) = - 3x - 1

3) (f+g) (3) = 18

Step-by-step explanation:

Given:

f (x) = x+1

g (x) = 4x + 2

1) To find (fg) (x)

(fg) (x) = f (x) * g (x)

= (x+1) (4x+2)

= (4x^2 + 2x+4x+2)

(fg) (x) = 4x^2 + 6x + 2

2) To find (f-g) (x)

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

= (x+1) - (4x+2)

(f-g) (x) = - 3x - 1

3) To find (f+g) (3)

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

(f+g) (3) = f (3) + g (3)

= (3+1) + (4 (3) + 2)

(f+g) (3) = 18