Question 419121
f(x) = x^3 - 7


set y = f(x) to get:


y = x^3 - 7


solve for x.


add 7 to both sides of the equation to get y + 7 = x^3


raise both sides of the equation to the power of (1/3) to get:


(y+7)^(1/3) = (x^3)^(1/3)


if a = b, then b = a, so you can flip the equation to get:


(x^3)^(1/3) = (y+7)^(1/3)


since (x^a)^b) = x^(a*b), your equation becomes:


x^(3*(1/3)) = (y+7)^(1/3) which becomes:


x = (y+7)^(1/3)


replace the x with y and the y with x to get:


y = (x+7)^(1/3)


that's your inverse equation.


let f^(-1)(x) = your new y and you get:


f^(-1)(x) = (x+7)^(1/3)


your original equation is f(x) = x^3 - 7


your inverse equation is f^(-1)(x) = (x+7)^(1/3


to prove it's an inverse, we take any value of x and solve for f(x).


if x = 9, then f(x) = (9)^3 - 7 = 722.


In our inverse equation, we let x = 722 which was f(x) in our original equation.


our inverse equation becomes (722 + 7)^(1/3 = 729^(1/3) = 9


x = 9 and f(x) = 722 in our original equation
x = 722 and f(-1)(x) = 9 in our inverse equation.


our inverse equation is good because it undoes what our original equation did.


the graph of both equations is shown below:


{{{graph(600,600,-20,20,-20,20,x^3-7,(x+7)^(1/3),x)}}}


I was not able to properly graph the equation of (x+7)^(1/3) when x is less than -7 using the built in graphing algorithm of algebra.com.


that piece is missing.


the following picture of another graph of the same equations shows you what it should look like.


<img src = "http://theo.x10hosting.com/problems/419121.jpg" alt = "***** picture not found *****" / >


since the graphs are inverse functions, they show up as reflections about the line y = x.