k(x) = (3g + 5h)(x) β (1)
k(x) = (5h - 3g)(x) β (3)
k(x) = (h - g)(x) β (2)
k(x) = (g + h)(x) β (4)
k(x) = (5g + 3h)(x) β (5)
k(x) = (3h - 5g)(x) β (6)
Step-by-step explanation:
* To solve this problem we will substitute h(x) and g(x) in k(x) in the
Β right column to find the corresponding function formula in the
Β left column
β΅ h(x) = 5 - 3x
β΅ g(x) = -3^x + 5
- Lets start with the right column
# k(x) = (3g + 5h)(x)
β΅ g(x) = -3^x + 5
β΅ 3g(x) = 3[-3^x + 5] = [3 Γ -3^x + 3 Γ 5]
- Lets simplify 3 Γ -3^x
Β take the negative out -(3 Γ 3^x), and use the rule a^n Γ a^m = a^(n+m)
β΄ -3(3 Γ 3^x) = -(3^x+1)
β΄ 3g(x) = -3^x+1 + 15
β΅ h(x) = 5 - 3x
β΅ 5h(x) = 5[5 - 3x] = [5 Γ 5 - 5 Γ 3x] = 25 - 15x
- Now substitute 3g(x) and 5h(x) in k(x)
β΅ k(x) = (3g + 5h)(x)
β΄ k(x) = -3^x+1 + 15 + 25 - 15x β simplify
β΄ k(x) = 40 - 3^x+1 - 15x
β΄ k(x) = 40 - 3^x+1 - 15x β k(x) = (3g + 5h)(x)
* k(x) = (3g + 5h)(x) β (1)
# k(x) = (5h - 3g)(x)
β΅ 5h(x) = 25 - 15x
β΅ 3g(x) = -3^x+1 + 15
β΅ k(x) = (5h - 3g)(x)
β΄ k(x) = 25 - 15x - (-3^x+1 + 15) = 25 -15x + 3^x+1 - 15 β simplify
β΄ k(x) = 10 + 3^x+1 - 15x
β΄ k(x) = 10 + 3^x+1 - 15x β k(x) = (5h - 3g)(x)
* k(x) = (5h - 3g)(x) β (3)
# k(x) = (h - g)(x)
β΅ h(x) = 5 - 3x
β΅ g(x) = -3^x + 5
β΅ k(x) = (h - g)(x)
β΄ k(x) = 5 - 3x - (-3^x + 5) = 5 - 3x + 3^x - 5 β simplify
β΄ k(x) = 3^x - 3x
β΄ k(x)= 3^x - 3x β k(x) = (h - g)(x)
* k(x) = (h - g)(x) β (2)
# k(x) = (g + h)(x)
β΅ h(x) = 5 - 3x
β΅ g(x) = -3^x + 5
β΅ k(x) = (g + h)(x)
β΄ k(x) = -3^x + 5 + 5 - 3x β simplify
β΄ k(x) = 10 - 3^x - 3x
β΄ k(x)= 10 - 3^x - 3x β k(x) = (g + h)(x)
* k(x) = (g + h)(x) β (4)
# k(x) = (5g + 3h)(x)
β΅ g(x) = -3^x + 5
β΅ 5g(x) = 5[-3^x + 5] = [5 Γ -3^x + 5 Γ 5] = 5(-3^x) + 25
β΄ 5g(x) = -5(3^x) + 25
β΅ h(x) = 5 - 3x
β΅ 3h(x) = 3[5 - 3x] = [3 Γ 5 - 3 Γ 3x] = 15 - 9x
- Now substitute 5g(x) and 3h(x) in k(x)
β΅ k(x) = (5g + 3h)(x)
β΄ k(x) = -5(3^x) + 25 + 15 - 9x β simplify
β΄ k(x) = 40 - 5(3^x) - 9x
β΄ k(x) = 40 - 5(3^x) - 9x β k(x) = (5g + 3h)(x)
* k(x) = (5g + 3h)(x) β (5)
# k(x) = (3h - 5g)(x)
β΅ 3h(x) = 15 - 9x
β΅ 5g(x) = -5(3^x) + 25
β΅ k(x) = (3h - 5g)(x)
β΄ k(x) = 15 - 9x - [-5(3^x) + 25] = 15 - 9x + 5(3^x) - 25 β simplify
β΄ k(x) = 5(3^x) - 9x - 10
β΄ k(x) = 5(3^x) - 9x - 10 β k(x) = (3h - 5g)(x)
* k(x) = (3h - 5g)(x) β (6)