Untuk g(x) = 2x² + 5x: [tex]\begin{aligned}(g\circ f)(x)&=g\left(f(x)\right)\\&=g\left(x^2+1\right)\\&=2\left(x^2+1\right)^2+5\left(x^2+1\right)\\&=2\left(x^4+2x^2+1\right)+5x^2+5\\&=2x^4+4x^2+2+5x^2+5\\(g\circ f)(x)&=\boxed{\,2x^4+9x^2+7\,}\end{aligned}[/tex]
Untuk g(x) = –2x² – 5x, karena fungsi g(x) ini merupakan negatif dari fungsi g(x) sebelumnya, (g\circ f)(x) juga merupakan negatif dari (g\circ f)(x) yang telah diperoleh sebelumnya, sehingga: [tex](g\circ f)(x)=\boxed{\,-2x^4-9x^2-7\,}[/tex]
[tex]\begin{aligned}\sf a.\ \ &(g\circ f)(x)=3x^2+24x+42\\\sf b.\ \ &(g\circ f)(x)=\begin{cases}2x^4+9x^2+7\\-2x^4-9x^2-7\end{cases}\\\sf c.\ \ &(g\circ f)(x)=\begin{cases}x^2-2\\-x^2+2\end{cases}\\\sf d.\ \ &(g\circ f)(x)=\begin{cases}2\sqrt{x^2+5}+3\\-2\sqrt{x^2+5}-3\end{cases}\end{aligned}[/tex]
Penjelasan dengan langkah-langkah:
Komposisi Fungsi
Untuk semua pertanyaan, yang ditanyakan adalah [tex](g\circ f)(x)[/tex].
Soal a
Diketahui:
[tex]\begin{aligned}\bullet\ &(f\circ g)(x)=3x^2-2\\\bullet\ &f(x)=x+4\end{aligned}[/tex]
Penyelesaian:
[tex]\begin{aligned}(f\circ g)(x)&=3x^2-2\\f\left(g(x)\right)&=3x^2-2\\g(x)+4&=3x^2-2\\\Rightarrow\ g(x)&=3x^2-6\\\end{aligned}[/tex]
Maka:
[tex]\begin{aligned}(g\circ f)(x)&=g\left(f(x)\right)\\&=g(x+4)\\&=3(x+4)^2-6\\&=3\left(x^2+8x+16\right)-6\\&=3x^2+24x+48-6\\(g\circ f)(x)&=\boxed{3x^2+24x+42}\end{aligned}[/tex]
______________
Soal b
Diketahui:
[tex]\begin{aligned}\bullet\ &(f\circ g)(x)=4x^4+20x^3+25x^2+1\\\bullet\ &f(x)=x^2+1\end{aligned}[/tex]
Penyelesaian:
[tex]\begin{aligned}(f\circ g)(x)&=4x^4+20x^3+25x^2+1\\f\left(g(x)\right)&=4x^4+20x^3+25x^2+1\\\left(g(x)\right)^2+1&=4x^4+20x^3+25x^2+1\\\left(g(x)\right)^2&=4x^4+20x^3+25x^2\\&=x^2\left(4x^2+20x+25\right)\\&=x^2(2x+5)^2\\&=\left(x(2x+5)\right)^2\\g(x)&=\pm\, x(2x+5)\\g(x)&=\begin{cases}x(2x+5)\\-x(2x+5)\end{cases}\\\Rightarrow\ g(x)&=\begin{cases}2x^2+5x\\-2x^2-5x\end{cases}\end{aligned}[/tex]
Maka:
[tex]\begin{aligned}(g\circ f)(x)&=g\left(f(x)\right)\\&=g\left(x^2+1\right)\\&=2\left(x^2+1\right)^2+5\left(x^2+1\right)\\&=2\left(x^4+2x^2+1\right)+5x^2+5\\&=2x^4+4x^2+2+5x^2+5\\(g\circ f)(x)&=\boxed{\,2x^4+9x^2+7\,}\end{aligned}[/tex]
[tex](g\circ f)(x)=\boxed{\,-2x^4-9x^2-7\,}[/tex]
Dengan demikian:
[tex](g\circ f)(x)=\begin{cases}2x^4+9x^2+7\\-2x^4-9x^2-7\end{cases}[/tex]
______________
Soal c
Diketahui:
[tex]\begin{aligned}\bullet\ &(f\circ g)(x)=x^2-4x+4\\\bullet\ &f(x)=x^2\\\bullet\ &g(3)\leftarrow ???\ \sf(tidak\ jelas)\end{aligned}[/tex]
Penyelesaian:
[tex]\begin{aligned}(f\circ g)(x)&=x^2-4x+4\\f\left(g(x)\right)&=(x-2)^2\\\left(g(x)\right)^2&=(x-2)^2\\g(x)&=\pm\,(x-2)\\g(x)&=\begin{cases}x-2\\-x+2\end{cases}\end{aligned}[/tex]
Maka:
[tex]\begin{aligned}(g\circ f)(x)&=g\left(f(x)\right)\\&=g(x^2)\\(g\circ f)(x)&=\boxed{\,x^2-2\,}\end{aligned}[/tex]
[tex]\begin{aligned}(g\circ f)(x)&=g\left(f(x)\right)\\&=g(x^2)\\(g\circ f)(x)&=\boxed{\,-x^2+2\,}\end{aligned}[/tex]
Dengan demikian:
[tex](g\circ f)(x)=\begin{cases}x^2-2\\-x^2+2\end{cases}[/tex]
______________
Soal d
Diketahui:
[tex]\begin{aligned}\bullet\ &(f\circ g)(x)=\sqrt{4x^2+12x+14}\\\bullet\ &f(x)=\sqrt{x^2+5}\\\bullet\ &g(2)=7\end{aligned}[/tex]
Penyelesaian:
[tex]\begin{aligned}(f\circ g)(x)&=\sqrt{4x^2+12x+14}\\f\left(g(x)\right)&=\sqrt{4x^2+12x+14}\\\sqrt{\left(g(x)\right)^2+5}&=\sqrt{4x^2+12x+14}\\\left(g(x)\right)^2+5&=4x^2+12x+14\\\left(g(x)\right)^2&=4x^2+12x+9\\\left(g(x)\right)^2&=(2x+3)^2\\g(x)&=\pm\,(2x+3)\\\Rightarrow\ g(x)&=\begin{cases}2x+3\\-2x-3\end{cases}\end{aligned}[/tex]
Maka:
[tex]\begin{aligned}(g\circ f)(x)&=g\left(f(x)\right)\\&=g\left(\sqrt{x^2+5}\right)\\(g\circ f)(x)&=\boxed{\,2\sqrt{x^2+5}+3\,}\end{aligned}[/tex]
[tex]\begin{aligned}(g\circ f)(x)&=g\left(f(x)\right)\\&=g\left(\sqrt{x^2+5}\right)\\(g\circ f)(x)&=\boxed{\,-2\sqrt{x^2+5}-3\,}\end{aligned}[/tex]
Dengan demikian:
[tex](g\circ f)(x)=\begin{cases}2\sqrt{x^2+5}+3\\-2\sqrt{x^2+5}-3\end{cases}[/tex]