Question

Drag the tiles to the correct boxes to complete the palrs.

Answer 1

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:

$f(x) = 2x + 3$

$g(x) = x - 1$

(f+g)(2)

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

At $x = 2$

$(f+g)(2) = 3(2) - 2 = 6 - 2 = 4$

Then

(f+g)(2)  = 4

(f-g)(4)

$(f-g)(x) = f(x) - g(x) = 2x + 3 - (x - 1) = 2x + 3 - x + 1 = x + 4$

At x = 4

$(f-g)(4) = 4 + 4 = 8$

Then

(f-g)(4) = 8

(f ÷g)(2)

$(f \div g)(x) = \frac{f(x)}{g(x)} = \frac{2x+3}{x-1}$

At x = 2

$(f \div g)(2) = \frac{7}{1} = 7$

Then

(f ÷g)(2) = 7

(f x g)(1)

$(f \times g)(x) = f(x)g(x) = (2x+3)(x-1) = 2x^2 -2x + 3x - 3 = 2x^2 + x - 3$

Then

$(f \times g)(1) = 2(1)^2 + 1 - 3 = 3 + 1 - 3 = 0$

So

(f x g)(1) = 0