Question 2095
Not too sure of your notation here, sorry...

inverse of a function is purely, re-write it as x=, that is all...

so, y={{{3x^3-1}}} is {{{x^3 = ((y+1)/(3))}}}

so x = cube root of {{{((y+1)/(3))}}}

so {{{f^(-1)(x)}}} = cube root of {{{((x+1)/(3))}}}


2. f(x) = {{{2x^2+1}}} and g(x)=x-4

Now, i am assuming that your fog(x) is just the same as fg(x)???. If so, fg(x) is f(x-4)...now just put this "x" value in the f(x) equation....

ie f(x-4) = {{{2(x-4)^2+1}}}

If you want to expand then do so...f(x-4) = 2(x^2-8x+16)+1 so this becomes

f(x-4) = 2x^2-16x+33.


jon