Question

tolong dijawab dengan benar dan jangan ngasal menggunakan cara​

Answer 1

[tex]\begin{aligned}\sf6.\,a.\ \ &f(x)=x+7\\\sf b.\ \ &f(x)=2x+1\end{aligned}[/tex]
 

Penjelasan dengan langkah-langkah:

Komposisi Fungsi

Nomor 6a

Diberikan fungsi:

[tex]\begin{aligned}g(x)&=x^2+5x-4\,,\ (f\circ g)(x)=x^2+5x+3\end{aligned}[/tex]

Kita akan menentukan fungsi f(x).

[tex]\begin{aligned}(f\circ g)(x)&=x^2+5x+3\\f\left ( g(x) \right )&=x^2+5x+3\\f\underbrace{\left(x^2+5x-4\right)}_{g(x)}&=\underbrace{\left(x^2+5x-4\right)}_{g(x)}+7\\f\left ( g(x) \right )&=g(x)+7\\\therefore\ f(x)&=x+7\end{aligned}[/tex]

__________________

Nomor 6b.

Diberikan fungsi:

[tex]\begin{aligned}g(x)&=x^2-3x+2\,,\ (f\circ g)(x)=2x^2-6x+5\end{aligned}[/tex]

Kita akan menentukan fungsi f(x).

[tex]\begin{aligned}(f\circ g)(x)&=2x^2-6x+5\\f\left ( g(x) \right )&=2x^2-6x+4+1\\f\underbrace{\left(x^2-3x+2\right)}_{g(x)}&=2\underbrace{\left(x^2-3x+2\right)}_{g(x)}+1\\f\left ( g(x) \right )&=2g(x)+1\\\therefore\ f(x)&=2x+1\end{aligned}[/tex]