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]
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
Verified answer
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]