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PLEASE HELP ASAP Due in 1 day!!!
Question 1
Find(f+g)(x)

f(x)=4x−4 and g(x)=2x^2−3x

Question 2
Find(f−g)(x)

f(x)=2x^2−2 and g(x)=4x+1

Question 3
Find(h⋅g)(x)

h(x)=3x−3 and g(x)=x^2+3

Question 4
Find(f/g)(x)and state the domain restriction.

f(x)=x+4 and g(x)=x+6

Respuesta :

Eyadaqrabawi

Step-by-step explanation:

Q1 . (f+g)(x) = f(x) + g(x)

=4x-4 +2x^2 -3x

= 2x^2 + x -4

Q2. (f-g)(x) = f(x) - g(x)

= 2x^2−2 - (4x+1)

= 2x^2 -2 -4x -1

= 2x^2 - 4x -3

Q3. h(x)=3x−3 and g(x)=x^2+3

(h.g)(x) = h(x) × g(x)

= (3x-3) × (x^2 + 3)

=3x^3 -3x^2 + 9x -9

Q4.f(x)=x+4 and g(x)=x+6

(f/g)(x) = f(x) ÷ g(x)

= x+4 / x+6

the domain restriction is x>-6

x<-6

x doesn't equal (-6)

If the two function f(x) and g(x) are to be combined, use basic arithmetic rules,

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

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

[tex](\frac{f}{g})(x)=\frac{f(x)}{g(x)}[/tex]

(f . g)(x) = f(x) × g(x)

Question 1

f(x) = 4x - 4 and g(x) = 2x² - 3x

[tex](f+g)(x)=f(x)+g(x)[/tex]

Therefore, [tex](f+g)(x)=(4x-4)+(2x^2-3x)[/tex]

                                   [tex]=2x^2+x-4[/tex]

Question 2

f(x) = 2x² - 2 and g(x) = 4x + 1

Since, (f - g)(x) = f(x) - g(x)

Therefore, (f - g)(x) = (2x² - 2) - (4x + 1)

                               = 2x² - 2 - 4x - 1

                               = 2x² - 4x - 3

Question 3

h(x) = 3x - 3 and g(x) = x² + 3

Since, (h - g)(x) = h(x) - g(x)

Therefore, (h - g)(x) = (3x - 3) - (x² + 3)

                                = 3x - 3 - x² - 3

                                = -x² + 3x - 6

Question 4

F(x) = x + 4 and g(x) = x + 6

Since, [tex](\frac{f}{g})(x)=\frac{f(x)}{g(x)}[/tex]

Therefore, [tex](\frac{f}{g})(x)=\frac{x+4}{x+6}[/tex]

Learn more,

https://brainly.com/question/7447002

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