Respuesta :
Answer:
(f+g)(2) = 4
(f-g)(4) = 8
(f ÷g)(2) = 7
(f x g)(1) = 0
Step-by-step explanation:
We are given these following functions:
[tex]f(x) = 2x + 3[/tex]
[tex]g(x) = x - 1[/tex]
(f+g)(2)
[tex](f+g)(x) = f(x) + g(x) = 2x + 3 + x - 1 = 3x - 2[/tex]
At [tex]x = 2[/tex]
[tex](f+g)(2) = 3(2) - 2 = 6 - 2 = 4[/tex]
Then
(f+g)(2) = 4
(f-g)(4)
[tex](f-g)(x) = f(x) - g(x) = 2x + 3 - (x - 1) = 2x + 3 - x + 1 = x + 4[/tex]
At x = 4
[tex](f-g)(4) = 4 + 4 = 8[/tex]
Then
(f-g)(4) = 8
(f ÷g)(2)
[tex](f \div g)(x) = \frac{f(x)}{g(x)} = \frac{2x+3}{x-1}[/tex]
At x = 2
[tex](f \div g)(2) = \frac{7}{1} = 7[/tex]
Then
(f ÷g)(2) = 7
(f x g)(1)
[tex](f \times g)(x) = f(x)g(x) = (2x+3)(x-1) = 2x^2 -2x + 3x - 3 = 2x^2 + x - 3[/tex]
Then
[tex](f \times g)(1) = 2(1)^2 + 1 - 3 = 3 + 1 - 3 = 0[/tex]
So
(f x g)(1) = 0
