Answer:
Remember, if f and g are functions then [tex](f\circ g)(x)=f(g(x))[/tex], then:
a)[tex](g\circ f)(x)=g(f(x)))=g(x^2)=2(x^2)+3=2x^2+3[/tex]
b) [tex]f(f(-2))=f((-2)^2)=f(4)=4^2=16[/tex]
c)
[tex](f \circ g)(\frac{1}{x})=f(g(\frac{1}{x}))\\=f(2(\frac{1}{x})+3)\\=f(\frac{2}{x}+3)\\=(\frac{2}{x}+3)^2\\ =(\frac{2}{x})^2+2*\frac{2}{x}*3+3^2 \\ =\frac{4}{x^2}+\frac{12}{x}+9[/tex]
then, [tex](f\circ g)=\frac{4}{x^2}+\frac{12}{x}+9[/tex]
