Question 113796
#3


{{{f(x)=x^3+2}}} Start with f(x)


{{{f(g(x))=(g(x))^3+2}}} Plug in {{{x=g(x)}}}



{{{f(g(x))=(root(3,x-2))^3+2}}} Replace each {{{g(x)}}} with {{{root(3,x-2)}}}. This is given since {{{g(x)=root(3,x-2)}}}



{{{f(g(x))=(x-2)+2}}} Cube the cube root (these two functions "undo" each other and cancel out)



{{{f(g(x))=x}}} Combine like terms



So this shows that {{{f(g(x))=x}}} 



{{{g(x)=root(3,x-2)}}} Start with g(x)



{{{g(f(x))=root(3,f(x)-2)}}}  Plug in {{{x=f(x)}}}



{{{g(f(x))=root(3,x^3+2-2)}}}  Replace f(x) with {{{x^3+2}}}



{{{g(f(x))=root(3,x^3)}}}  Combine like terms



{{{g(f(x))=x}}}  Take the cube root of {{{x^3}}} to get x. Remember the cube root and the cube cancel each other out.





So this shows that {{{g(f(x))=x}}} 



Now if we graph the two functions, we get



{{{ graph( 500, 500, -10, 10, -10, 10,x^3+2, (x-2)^(1/3),-(-x+2)^(1/3)) }}}