#FinishLine (GearShark Book 5)

#FinishLine (GearShark Book 5)

Matematika Jika f(x) = 2x+1 dan g∘fx=4x2+2 maka g-127=… a. 6 b. 5 c. 27 d. 26 e. 25 ​

Jika f(x) = 2x+1 dan g∘fx=4x2+2 maka g-127=… a. 6 b. 5 c. 27 d. 26 e. 25 ​

Penjelasan dengan langkah-langkah:

[tex]f(x) = 2x + 1[/tex]

[tex] \: [/tex]

Misal

[tex]2a + 1 = x[/tex]

[tex]2a = x - 1[/tex]

[tex]a = \frac{x - 1}{2} [/tex]

[tex] \: [/tex]

Maka :

[tex]g(x) = 4 {x}^{2} + 2[/tex]

[tex]g(x) = 4 {( \frac{x - 1}{2} )}^{2} + 2 [/tex]

[tex]g(x) = 4. \frac{ {(x - 1)}^{2} }{4} + 2[/tex]

[tex]g(x) = {x}^{2} - 2x + 1 + 2[/tex]

[tex]g(x) = {x}^{2} - 2x + 3[/tex]

[tex] \: [/tex]

—Cari invers dan x = 27

[tex]y = {x}^{2} - 2x + 3[/tex]

[tex] {y}^{2} - 2y + 3= x[/tex]

[tex] {y}^{2} - 2y + 3 - 2 = x - 2[/tex]

[tex] {y}^{2} - 2y + 1 = x - 2[/tex]

[tex] {(y - 1)}^{2} = x - 2[/tex]

[tex]y - 1 = \pm \sqrt{x - 2} [/tex]

[tex]y = 1 \pm \sqrt{x - 2} [/tex]

[tex] {g}^{ - 1} (27) = 1 \pm \sqrt{27 - 2} [/tex]

[tex]{g}^{ - 1} (27) = 1 \pm \sqrt{25} [/tex]

[tex]{g}^{ - 1} (27) = 1 \pm5[/tex]

[tex]{g}^{ - 1} (27) = 1 - 5 \: \text{atau} \: 1 + 5[/tex]

[tex]{g}^{ - 1} (27) = - 4 \: \text{atau} \: 6[/tex]

Salah satu jawaban ada 6. Jadi, jawabannya adalah A. 6

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