SOLUTION: Click on the link to solve this problem: I'm doing the review section in my calculus class, but it's been over two years since i took any math... The problem is to find the inv

Algebra ->  Inverses -> SOLUTION: Click on the link to solve this problem: I'm doing the review section in my calculus class, but it's been over two years since i took any math... The problem is to find the inv      Log On


   

Question 9258: Click on the link to solve this problem:
I'm doing the review section in my calculus class, but it's been over two years since i took any math...
The problem is to find the inverse of an equation, input a given number and solve. However, the review section on inverse only give simple equations, the ones in the homework are more complex..
find the inverse of:
f%28x%29=x%5E3%2B2x-1
So I made f(x) into "y" and I exhanged it with x so it now looks like
x=y%5E3%2B2y-1
I start solving for y
x%2B1=y%5E3%2B2y
This is where I start to get stuck I've factored, but it still leaves a y
x%2B1=y%28y%5E2%2B2%29
or
I've tried to cube root it, but then I still have a y under the radican
x%2B1%5E%281%2F3%29=%28y%5E3%2B2y-1%29%5E%281%2F3%29
I'm stuck..
help please,
thank you
Answer by ichudov(507) About Me  (Show Source):
You can put this solution on YOUR website!
The hard core way of using this would be to simply use the formula for roots of a cubic equation. Check out this article and follow the formula. It becomes easier if the square component is missing, as it is in your case. I cannot see any easier solution. I wonder if you misinterpret the question that was asked. What was the text of the question exactly?