haiii kakk mohon bantuan dan cara mengerjakannyaaa
Answer (1)
[tex]\rm{\bf{Penyelesaian}}[/tex]1. [tex](f \circ g)(x) = f(g(x)) = 4x - 3[/tex][tex]f(g(x)) = 2g(x) + 1 = 4x - 3[/tex][tex]2g(x) = 4x - 3 - 1[/tex][tex]2g(x) = 4x - 4[/tex][tex]g(x) = \frac{4x - 4}{2}[/tex][tex]g(x) = 2x - 2[/tex]2. [tex](g \circ f)(x) = g(f(x)) = x^2 + 6x + 9[/tex][tex]g(x + 3) = x^2 + 6x + 9[/tex][tex]g(y) = (y - 3)^2 + 6(y - 3) + 9[/tex][tex]g(y) = (y^2 - 6y + 9) + 6y - 18 + 9[/tex][tex]= y^2 - 6y + 9 + 6y - 18 + 9[/tex][tex]= y^2 + 0y + 0[/tex][tex]= y^2[/tex][tex]g(x) = x^2[/tex]3. [tex](f \circ g)(x) = f(g(x)) = \sqrt{4x + 8}[/tex][tex]g(x) = 2x + 4[/tex][tex]f(2x + 4) = \sqrt{4x + 8}[/tex][tex]f(y) = \sqrt{4 \left( \frac{y - 4}{2} \right) + 8}[/tex][tex]= \sqrt{2(y - 4) + 8}[/tex][tex]= \sqrt{2y - 8 + 8}[/tex][tex]= \sqrt{2y}[/tex][tex]f(x) = \sqrt{2x}[/tex]4. [tex](g \circ f)(x) = g(f(x)) = \frac{1}{x + 2}[/tex][tex]g(x - 1) = \frac{1}{x + 2}[/tex][tex]g(y) = \frac{1}{(y + 1) + 2} = \frac{1}{y + 3}[/tex][tex]g(x) = \frac{1}{x + 3}[/tex]
