tolong di jawab semoga semoga dibalas oleh allah

No.2

f(x) = 3x + 1  ;  x < 0
         x² + 1   ;  0 ≤ x < 2
         2x + 3  ;  x ≥ 2
maka
f(-1) + f(1) + f(3)
= [3(-1) + 1] + [1² + 1] + [2(3) + 3]
= -2 + 2 + 9
= 9

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No. 3

f(x) = 2x² - 2x + 14
g(x) = x² + x

fog(x)
= f(g(x))
= f(x² + x)
= 2(x² + x)² - 2(x² + x) + 14
= 2(x⁴ + 2x³ + x²) - 2x² - 2x + 14
= 2x⁴ + 4x³ + 2x² - 2x² - 2x + 14
= 2x⁴ + 4x³ - 2x + 14

fog(-1)
= 2(-1)⁴ + 4(-1)³ - 2(-1) + 14
= 2 - 4 + 2 + 14
= 14

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No. 4
A.
f(x) = 24x + 12
y = 24x + 12
y - 12 = 24x
(y - 12)/24 = x
jadi
f⁻¹(x) = (x - 12) / 24

B.
f(x) = (9x - 5) / 12x
y = (9x - 5) / 12x
12xy = (9x - 5)
12xy - 9x = -5
x (12y - 9) = -5
x = -5 / (12y - 9)
jadi 
f⁻¹(x) = -5 / (12x - 9)

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No.5
A.
f(x)= 2x + 7
y = 2x + 7
y - 7 = 2x
(y - 7) / 2 = x
jadi
f⁻¹(x) = (x - 7) / 2

B.
g(x)= 3 / 2x
y = 3 / 2x
2xy = 3
x = 3/2y
jadi 
g⁻¹(x) = 3 / 2x