Jika diketahui f(x) = x – 3 dan g(x) = 2x + 4 maka (g o f)^-1(2) adalah …
Jika f(x) = x² – 4 maka f^-1(x) adalah…
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gunakan cara ya !!!
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Penjelasan dengan langkah-langkah:
[tex]f(x) = x - 3[/tex]
[tex]g(x) = 2x + 4[/tex]
[tex] \: [/tex]
[tex](g \circ f)(x) = g(f(x))[/tex]
[tex](g \circ f)(x) = 2(x - 3) + 4[/tex]
[tex](g \circ f)(x) = 2x - 6 + 4[/tex]
[tex](g \circ f)(x) = 2x - 2[/tex]
[tex]y = 2x - 2[/tex]
[tex]2y - 2 = x[/tex]
[tex]2y = x + 2[/tex]
[tex]y = \frac{x + 2}{2} [/tex]
[tex] {(g \circ f)}^{ - 1} (x) = \frac{x + 2}{2} [/tex]
[tex](g \circ f) {}^{ - 1} (2) = \frac{2 + 2}{2} = \frac{4}{2} = 2[/tex]
[tex] \: [/tex]
[tex] \: [/tex]
[tex]f(x) = {x}^{2} - 4[/tex]
[tex]y = {x}^{2} - 4[/tex]
[tex] {y}^{2} - 4 = x[/tex]
[tex] {y}^{2} = x + 4[/tex]
[tex]y = \pm \sqrt{x + 4} [/tex]
[tex] {f}^{ - 1} (x) = \pm \sqrt{x + 4} [/tex]
Penjelasan dengan langkah-langkah:
f(x)=x-3, g(x)=2x+4
(gof)^-1(2)?
g{f(x)}=2(x-3)+4=2x-6+4=2x-2, misal (gof)(x)=n
2x-2=(gof)(x)
2x-2=n
2x=n+2 → x=(n+2)/2
(gof)^-1(x)=(n+2)/2=(x+2)/2=½(x+2)
(gof)^-1(2)=½(2+2)=½(4)=2
.
f(x)=x^2 -4 , misal f(x)=n
x^2-4=f(x)
x^2-4=n
x^2=n+4
x=±√n+4
f^-1(x)=±√(n+4)=±√(x+4) → √(x+4) atau -√(n+4)
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