QUESTION 1
[tex] \boxed {f(x) = 2x + 6 \to \: g(x) = \frac{1}{2}x - 3 } [/tex]
The reason is that:
[tex]g(f(x)) = 2( \frac{1}{2} x - 3) + 6[/tex]
Expand:
[tex]g(f(x)) = x - 6+ 6 [/tex]
[tex]g(f(x)) = x[/tex]
QUESTION 2
[tex]\boxed {f(x) =3 - 2x \to \: g(x) = - \frac{1}{2}(x - 3)} [/tex]
The reason that
[tex]g(f(x)) = - \frac{1}{2} (3 - 2x - 3)[/tex]
[tex]g(f(x)) = - \frac{1}{2} (- 2x )[/tex]
[tex]g(f(x)) =x[/tex]
QUESTION 3
[tex]\boxed {f(x) = \sqrt[3]{3x}+ 2 \to \: g(x) = \frac{ {(x - 3)}^{3} }{3} } [/tex]
The reason is that:
[tex]f(g(x)) = \sqrt[3]{ \frac{3 {(x - 2)}^{3} }{3} } + 2[/tex]
[tex]f(g(x)) = x-2 + 2[/tex]
[tex]f(g(x))=x[/tex]
QUESTION 4
[tex]\boxed {f(x)=3\sqrt[3]{x + 2} \to \: g(x) = \frac{1}{27} {x}^{3} - 2} [/tex]
The reason is that
[tex]f(g(x)) = 3 \sqrt[3]{ \frac{1}{27} {x}^{3} - 2 + 2} [/tex]
[tex]f(g(x)) = 3 \sqrt[3]{ \frac{1}{27} {x}^{3} } [/tex]
[tex]f(g(x)) = 3 \times \frac{1}{3} x[/tex]
[tex]f(g(x)) =x[/tex]
