Question

f(x)= 2x + 1/ x-3 , g(x) = 4-2x/ x+2 (Fog)(x) = (gof)(x) =. tolong jawab dengan teliti plis jangan salah soal​

Answer 1

Jawab:

[tex]\begin{aligned}&{\sf{(f\circ{g})(x)=\frac{3x-10}{5x+2}}}\\&{\sf{(g\circ{f})(x)=-\frac{14}{4x-5}}}\end{aligned}[/tex]

Penjelasan dengan langkah-langkah:

[tex]\begin{aligned}{\sf{(f\circ{g})(x)}}&={\sf{f(g(x))}}\\&={\sf{\frac{2\left(\frac{4-2x}{x+2}\right)+1}{\frac{4-2x}{x+2}-3}}}\\&={\sf{\frac{\frac{2(4-2x)+(x+2)}{\cancel{x+2}}}{\frac{(4-2x)-3(x+2)}{\cancel{x+2}}}}}\\&={\sf{\frac{(8-4x)+(x+2)}{(4-2x)-(3x+6)}}}\\&={\sf{\frac{-4x+x+2+8}{-2x-3x-6-4}}}\\&={\sf{\frac{-3x-10}{-5x+2}}}\\&={\sf{-\left(-\frac{3x-10}{5x+2}\right)}}\\&={\sf{\frac{3x-10}{5x+2}}}\end{aligned}[/tex]

[tex]\begin{aligned}{\sf{(g\circ{f})(x)}}&={\sf{g(f(x))}}\\&={\sf{\frac{4-2\left(\frac{2x+1}{x-3}\right)}{\frac{2x+1}{x-3}+2}}}\\&={\sf{\frac{\frac{4(x-3)-2(2x+1)}{\cancel{x-3}}}{\frac{(2x+1)+2(x-3)}{\cancel{x-3}}}}}\\&={\sf{\frac{(4x-12)-(4x+2)}{(2x+1)+(2x-6)}}}\\&={\sf{\frac{4x-4x-2-12}{2x+2x-6+1}}}\\&={\sf{-\frac{14}{4x-5}}}\end{aligned}[/tex]