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)

omg what happend to your question? if you could edit this question i can

step-by-step explanation:

the solution is x=1

step-by-step explanation:

we are given our   equation as

let's assume

left side function is

right side function is

now, we can draw graph

and then we can get intersection point

the intersection point will be solution

so, the solution is

x=1