Question

For the given function f and g, complete parts (a)-(h). For parts (a)-(d), also find the domain.

Answer 1

Answer:

a) (f + g)(x) = 9x + 1

Domain: x ε R

b) (f - g)(x) = (-5x + 13)

Domain: x ε R

c) (f.g)(x) = 14x² + 37x - 42

Domain: x ε R

d) (f/g)(x) = (2x + 7)/(7x -6)

Domain: (x ε R except x = 6/7)

e) (f + g)(7) = 64

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

g) (f.g)(3) = 195

h) (f/g)(x) = 1/3

Step-by-step explanation:

f(x) = 2x + 7, g(x) = 7x - 6

a) (f + g)(x) = f(x) + g(x) = (2x + 7) + (7x - 6)

(f + g)(x) = 9x + 1

Since x is defined for functions f & g for all real numbers, the domain of (f + g)(x) is x ε R

b) (f - g)(x) = f(x) - g(x) = (2x + 7) - (7x - 6)

(f - g)(x) = (-5x + 13)

Since x is defined for functions f & g for all real numbers, the domain of (f - g)(x) is x ε R

c) (f.g)(x) = f(x) × g(x) = (2x + 7)(7x - 6)

(f.g)(x) = 14x² - 12x + 49x - 42 = 14x² + 37x - 42

Since x is defined for functions f & g for all real numbers, the domain of (f.g)(x) is x ε R

d) (f/g)(x) = f(x)/g(x) = (2x + 7)/(7x -6)

x is defined for functions f & g for all real numbers, the domain of (f/g)(x) will be x ε R except when the denominator vanishes (that is, goes to zero). This will cause the function to take up values of ∞.

This will happen when 7x - 6 = 0, x = 6/7.

Therefore, the domain of (f/g)(x) is x ε R except the point, x = 6/7.

e) (f + g)(7) = 9(7) + 1 = 64

f) (f - g)(2) = -5(2) + 13 = 3

g) (f.g)(3) = 14(3²) + 37(3) - 42 = 195

h) (f/g)(27) = (2(27) + 7)/(7(27) - 6) = 61/183 = 1/3

Hope this Helps!!!