Math / AlgebraQuestionFind (f∘g)(x)(f \circ g)(x)(f∘g)(x), (g∘f)(x)(g \circ f)(x)(g∘f)(x), and (f∘g)(3)(f \circ g)(3)(f∘g)(3) for f(x)=2x+1f(x)=2x+1f(x)=2x+1 and g(x)=x+34+xg(x)=\frac{x+3}{4+x}g(x)=4+xx+3.Studdy Solutionimplify the expression.(f∘g)()=127+=197(f \circ g)() = \frac{12}{7} + = \frac{19}{7}(f∘g)()=712+=719So, the solutions area. (f∘g)(x)=2(x+)4+x+(f \circ g)(x) = \frac{2(x+)}{4+x} +(f∘g)(x)=4+x2(x+)+ b. (g∘f)(x)=2x+42x+5(g \circ f)(x) = \frac{2x+4}{2x+5}(g∘f)(x)=2x+52x+4 c. (f∘g)()=197(f \circ g)() = \frac{19}{7}(f∘g)()=719 View Full Solution - FreeWas this helpful?
