Jawab:
(f·g·h)(x) = 2x²-4x-5
Ciri-ciri fungsi (f·g·h)(x) =

• Domain -∞ < x < ∞
• Range f(x) ≥ -7
• Titik balik = (1, -7)
• Titik potong sumbu y = (0, -5)
• Titik ptg sumbu x =
  (1 + ½(√14), 0) dan (1 - ½(√14), 0)

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Penjelasan dengan langkah-langkah:
Diketahui:
--> f(x) = 2x+1
--> g(x) = x²-4
--> h(x) = x-1
Tentukan (f·g·h)
Masukkan h(x) kedalam g(x) didalam f(x)
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(f·g·h)(x) = 2[[x-1]²-4]+1
(f·g·h)(x) = 2[[x-1][x-1]-4]+1
(f·g·h)(x) = 2[x²-2x+1-4]+1
(f·g·h)(x) = 2[x²-2x-3]+1
(f·g·h)(x) = 2x²-4x-6+1
(f·g·h)(x) = 2x²-4x-5
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Domain -∞ < x < ∞
fungsi kuadrat domain tak hingga
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Range f(x) ≥ -((b²-4ac)/(4a))
--> a = 2, b = -4, c = -5
Range f(x) ≥ -(((-4)²-4(2)(-5))/(4(2)))
Range f(x) ≥ -((16-(-40))/(8))
Range f(x) ≥ -(56/8)
Range f(x) ≥ -7
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Titik balik = (-b/2a, -((b²-4ac)/(4a)))
--> a = 2, b = -4, c = -5
Titik balik = (-(-4)/(2(2)), -(((-4)²-4(2)(-5))/(4(2))))
Titik balik = (4/4, -((16-(-40))/(8)))
Titik balik = (1, -(56/8))
Titik balik = (1, -7)
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Titik potong sumbu y, x = 0
y = 2x²-4x-5
y = 2(0)²-4(0)-5
y = -5
Titik potong sumbu y = (0, -5)
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Titik potong sumbu x, y = 0
2x²-4x-5 = 0
--> a = 2, b = -4, c = -5
x = -b±√(b²-4ac)/2a
x = -(-4)±√((-4)²-4(2)(-5))/(2(2))
x = 4±√(16-(-40))/(4)
x = 4/4 ± √(16+40)/(4)
x = 4/4 ± √(56)/(4)
x = 1 ± (√4√(14)/(4))
x = 1 ± (2√(14)/(4))
x = 1 ± (√14/2)
x = 1 + ½√14, dan 1 - ½√14
Titik ptg sumbu x =
(1 + ½(√14), 0) dan (1 - ½(√14), 0)
<(7o7)>