Matematika Sekolah Menengah Pertama 1. Jika,[tex]f(x) = \frac{x + 3}{2 - 6x} [/tex] X≠ -⅓ maka rumus fungsi[tex] {f}^{ - 1} (x)[/tex]adalah...2. Diket: fungsi F:R→R dirumuskan[tex]f(x) = \frac{6x + 8}{x} [/tex]x≠0. Agar [tex] {f}^{ - 1} (t) = 2[/tex]maka nilai t=​

yeseniaullrichulv – April 11, 2023

1. Jika,
[tex]f(x) = \frac{x + 3}{2 - 6x} [/tex]
X≠ -⅓
maka rumus fungsi
[tex] {f}^{ - 1} (x)[/tex]
adalah...


2. Diket: fungsi F:R→R dirumuskan
[tex]f(x) = \frac{6x + 8}{x} [/tex]
x≠0. Agar
[tex] {f}^{ - 1} (t) = 2[/tex]
maka nilai t=
​

Jawab:

Penjelasan dengan langkah-langkah:

1. f(x) = (x + 3)/(2 - 6x)

y = (x + 3)/(2 - 6x)

y (2 - 6x) = x + 3

-6xy + 2y = x + 3

2y - 3 = 6xy + x

2y - 3 = x(6y + 1)

x = (2y - 3)/(6y + 1)

f^-¹ (x) = (2x - 3)/(6x + 1)

x ≠ -⅙.

2. f(x) = (6x + 8)/x

y = (6x + 8)/x

xy = 6x + 8

xy - 6x = 8

x(y - 6) = 8

x = 8/(y - 6)

f^-¹ (x) = 8/(x - 6)

x ≠ 6

f^-¹ (t) = 8/(t - 6)

8/(t - 6) = 2

8 = 2(t - 6)

4 = t - 6

t = 4 + 6

t = 10

[answer.2.content]