Match each function with the corresponding function formula when h(x) = 2 – 2x and g(x) = –2 x + 2.

1 -> f

2 -> a

3 -> d

4 -> b

5 -> e

6 -> c

Step-by-step explanation:

We are given:

h(x) = 2 - 2x and g(x) = -2^x+2

a) k(x)= (g-h)(x) = g(x) - h(x)

              = -2^x+2 - (2 - 2x)

              = -2^x+2 - 2 + 2x

              = -2^x+2x or 2x-2^x

k(x) = 2x-2^x = (g-h)(x)

So, 2 matches with a

b) k(x)= (g+h)(x) = g(x) + h(x)

              = -2^x+2 + (2 - 2x)

              = -2^x+2 + 2 - 2x

              = -2^x+4-2x or 4 - 2^x -2x

So, k(x) = 4 - 2^x -2x = (g+h)(x)

So, 4 matches with b

c) k(x)= (2h - 2g)(x) = 2h(x) - 2g(x)

              =  2(2 - 2x)-2(-2^x+2)

              = 4 - 4x+2.2^x-4

              = -4x+2^x+1 or 2^x+1-4x

So, k(x) = 2^x+1-4x = (2h - 2g)(x)

So, 6 matches with c

d) k(x) = (h-2g)(x) = h(x) - 2g(x)

         = (2 - 2x)-2(-2^x+2)

         = 2 - 2x+2.2^x-4

         = -2-2x+2^x+1

         = 2^x+1-2x-2

So, k(x) = 2^x+1-2x-2 = (h-2g)(x)

So, 3 matches with d

e) k(x) = (h-g)(x) = h(x) - g(x)

          = (2 - 2x)-(-2^x+2)

              = 2 - 2x+2^x-2

              = -2x+2^x or 2^x-2x

So, k(x) =  2^x-2x = (h-g)(x)

So, 5 matches with e

f) k(x) = (2g+h)(x) = 2g(x)+h(x)

          = 2(-2^x+2) + (2 - 2x)

          = -2.2^x+4 + 2 - 2x

          = -2^x+1+6-2x or 6 - 2^x+1-2x

So, k(x) = 6 - 2^x+1-2x = (2g+h)(x)

So, 1 matches with f

1) 6 - 2^(x + 1) - 2x = (2g - h)(x)

2) 2x - 2^x = (g - h)(x)

3) 2^(x + 1) - 2x - 2 = (h - 2g)(x)

4) 4 - 2^x - 2x = (g + h)(x)

5) 2^x - 2x = (h - g)(x)

6) 2^(x + 1) - 4x = (2h - 2g)(x)

Step-by-step explanation:

* Lets use right column to find the corresponding function in the

 left column

∵ h(x) = 2 - 2x

∵ g(x) = -2^x + 2

# k(x) = (g - h)(x)

∵ k(x) = g(x) - h(x)

∴ k(x) = -2^x + 2 - (2 - 2x) ⇒ multiply the -ve sign by the bracket

- Remember -ve × +ve = -ve and -ve × -ve = +ve

∴ k(x) = -2^x + 2 - 2 + 2x

∴ k(x) = -2^x + 2x ⇒ (2)

∴ k(x) = (g - h)(x) ⇒ (2)

* 2x - 2^x = (g - h)(x)

# k(x) = (g + h)(x)

∵ k(x) = g(x) + h(x)

∴ k(x) = -2^x + 2 + (2 - 2x) ⇒ simplify

∴ k(x) = -2^x + 2 + 2 - 2x

∴ k(x) = -2^x + 4 - 2x ⇒ (4)

∴ k(x) = (g + h)(x) ⇒ (4)

* 4 - 2^x - 2x = (g + h)(x)

# k(x) = (2h - 2g)(x)

∵ k(x) = 2h(x) - 2g(x)

- Lets find 2h(x) and 2g(x)

∵ 2h(x) = 2[2 - 2x] = 2(2) - 2(2x) = 4 - 4x

∵ 2g(x) = 2[-2^x + 2] = 2(-2^x) + 2(2)

- Remember a^n × a^m = a^(n + m)

∴ 2g(x) = 2(-2^x) + 4 = - (2 × 2^x) + 4 = - 2^(x + 1) + 4

∴ k(x) = 4 - 4x - [-2^(x + 1) + 4] ⇒ simplify

∴ k(x) = 4 - 4x + 2^(x + 1) - 4

∴ k(x) = 2^(x + 1) - 4x ⇒ (6)

∴ k(x) = (2h - 2g)(x) ⇒ (6)

* 2^(x + 1) - 4x = (2h - 2g)(x)

# k(x) = (h - 2g)(x)

∵ k(x) = h(x) - 2g(x)

∵ h(x) = 2 - 2x =

∴ 2g(x) = - 2^(x + 1) + 4

∴ k(x) = 2 - 2x - [-2^(x + 1) + 4] ⇒ simplify

∴ k(x) = 2 - 2x + 2^(x + 1) - 4

∴ k(x) = 2^(x + 1) - 2x - 2 ⇒ (3)

∴ k(x) = (h - 2g)(x) ⇒ (3)

* 2^(x + 1) - 2x - 2 = (h - 2g)(x)

# k(x) = (h - g)(x)

∵ k(x) = h(x) - g(x)

∴ k(x) = 2 - 2x - [-2^x + 2] ⇒ multiply the -ve sign by the bracket

- Remember -ve × +ve = -ve and -ve × -ve = +ve

∴ k(x) = 2 - 2x + 2^x - 2

∴ k(x) = 2^x - 2x ⇒ (5)

∴ k(x) = (h - g)(x) ⇒ (5)

* 2^x - 2x = (h - g)(x)

# k(x) = (2g + h)(x)

∵ k(x) = 2g(x) + h(x)

∵ h(x) = 2 - 2x =

∴ 2g(x) = - 2^(x + 1) + 4

∴ k(x) = -2^(x + 1) + 4 + 2 - 2x ⇒ simplify

∴ k(x) = -2^(x + 1) - 2x + 6

∴ k(x) = 6 - 2^(x + 1) - 2x ⇒ (1)

∴ k(x) = (2g + h)(x) ⇒ (1)

* 6 - 2^(x + 1) - 2x = (2g - h)(x)

answer: it is d

step-by-step explanation: