Question

kalo bisa semuanya.....​

Answer 1

Verified answer

Diketahui [tex]f(x) = \dfrac{3}{x-2} [/tex] dan [tex] g(x) = \sqrt{x-4} [/tex].

a. Tentukan [tex] (f\circ g)(x) [/tex]

[tex] \begin{align} (f\circ g)(x) &= f(g(x)) \\ &= f(\sqrt{x-4}) \\ &= \dfrac{3}{\sqrt{x-4}-2} \blue{\cdot\frac{\sqrt{x-4}+2}{\sqrt{x-4}+2}} \\ &= \dfrac{3({\sqrt{x-4}+2})}{(x-4)-4} \\ &= \dfrac{3\sqrt{x-4}+6}{x-8}\end{align} [/tex]

b. Tentukan [tex] (g\circ f)(x) [/tex]

[tex] \begin{align} (g\circ f)(x) &= g(f(x)) \\ &= g\left(\frac{3}{x-2}\right) \\ &= \sqrt{\dfrac{3}{x-2}-4} \\ &= \sqrt{\dfrac{3-4(x-2)}{x-2}} \\ &= \sqrt{\dfrac{5-4x}{x-2}} \end{align} [/tex]

c. Tentukan [tex] D_{f\circ g} [/tex]

[tex] \begin{align} x-4 &≥ 0 \:\:\text{ dan } & x-8 &≠ 0 \\ x &\geq 4 & x &≠ 8 \end{align} [/tex]

Maka, [tex] D_{f\circ g} =\{x|x\geq 4, x≠8, x\in\R\} [/tex]

d. Tentukan [tex] D_{g\circ f} [/tex]

[tex] \begin{align} \dfrac{5-4x}{x-2} &\geq 0 \:\:\:\:\:\text{ dan}& x-2 &≠0 \\ \dfrac{4x-5}{x-2} &\leq 0 &x &≠2 \\ \dfrac{5}{4} \leq x &\leq 2\end{align} [/tex]

Maka, gabungan dari dua syarat diatas menghasilkan:

[tex] D_{f\circ g} =\left\{x|\tfrac{5}{4}\leq x<2, x\in\R\right\} [/tex]

e. Tentukan [tex] R_{f\circ g} [/tex]

[tex] R_{f\circ g} = \left\{ y| y≤ -\dfrac{3}{2}, y>0, y\in\R\right\} [/tex]

f. Tentukan [tex] R_{g\circ f} [/tex]

Karena fungsi nya merupakan akar pecahan, maka: [tex]R_{g\circ f} = \{y|y≥0,y\in\R\} [/tex]

Answer 2

Materi : Fungsi dan Relasi, Fungsi Komposisi

f(x) = 3/( x - 2 )

g(x) = √( x - 4 )

Bagian A

( f • g )(x) = f(g(x)) = f( √[ x - 4 ] )

= 3/( √[ x - 4 ] - 2 )

Bagian B

( g • f )(x) = g(f(x)) = g( 3/[ x - 2 ] )

= √( 3/[ x - 2 ] - 4 )

= √( 3/[ x - 2 ] - [ 4x - 8 ]/[ x - 2 ] )

= √( [ - 4x + 11 ]/[ x - 2 ] )

Bagian C

D( f • g ) = { x | x ≠ 8, x ∈ R }

Bagian D

D( g • f ) = { x | x ≠ 2, x ∈ R }

Bagian E

y = 3/( √[ x - 4 ] - 2 )

y( √[ x - 4 ] - 2 ) = 3

√[ x - 4 ] - 2 = 3/y

√[ x - 4 ] = 2y/y + 3/y

√[ x - 4 ] = ( 2y + 3 )/y

x - 4 = [ ( 2y + 3 )/y ]²

x - 4 = ( 4y² + 12y + 9 )/y²

x = 4y²/y² + ( 4y² + 12y + 9 )/y²

x = ( 8y² + 12y + 9 )/y²

R( f • g ) = { y | y ≠ 0 }

Bagian F

y = √( [ - 4x + 11 ]/[ x - 2 ]

y² = ( - 4x + 11 )/( x - 2 )

y²( x - 2 ) = - 4x + 11

xy² - 2y² = - 4x + 11

xy² + 4x = 2y² + 11

x( y² + 4 ) = 2y² + 11

x = ( 2y² + 11 )/( y² + 4 )

R( g • f ) = { y | y ≠ i√4 = √(-4) }

Semoga bisa membantu

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