Troubleshooting Inverse Functions: Solving Algebraic Problems

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SUMMARY

The discussion centers on troubleshooting the inverse function problem defined by the equation y = cuberoot(x+sqrt(1+x^2)) + cuberoot(x-sqrt(1+x^2). The user initially attempted to solve for y by cubing both sides and simplifying, leading to an incorrect conclusion of y = (1/2)x^3. However, the correct approach involves recognizing that (a+b)³ = a³ + b³ + 3ab(a+b), which allows for the proper manipulation of the equation to arrive at the correct answer, y = (1/2)(3x+x^3).

PREREQUISITES
  • Understanding of inverse functions and their properties
  • Familiarity with algebraic manipulation techniques
  • Knowledge of cube roots and their properties
  • Ability to apply the binomial theorem in algebraic expressions
NEXT STEPS
  • Study the properties of inverse functions in algebra
  • Learn advanced algebraic manipulation techniques, including the binomial theorem
  • Practice solving equations involving cube roots and square roots
  • Explore common mistakes in algebraic problem-solving and how to avoid them
USEFUL FOR

Students revisiting algebra concepts, educators teaching inverse functions, and anyone seeking to improve their problem-solving skills in algebra.

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Ok, I decided to review basic algebra since I haven't done anything with it in like, forever. I came across an inverse function problem that I can't get the right answer.

the equation is:

y = cuberoot(x+sqrt(1+x^2)) + cuberoot(x-sqrt(1+x^2))

I tried replacing X with Y, and solving for Y
and getting rid of the cube roots by cubing both sides
X^3 = y + sqrt(1+y^2) + y - sqrt(1+y^2)
simplifying a bit (the square roots go away)
x^3 = 2y
so
y = (1/2)x^3

Yet the book I'm using says the answer is y=(1/2)(3x+x^3)

What did I do wrong?
 
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(a+b)³ =/= a³ + b³, as you have done.

Use (a+b)³ = a³ + b³ +3ab(a+b).

Start with y = a+b, and cube both sides. If you put a+b=y again, the result comes quickly,
 
thanks, I think I got it. Man, a few years ago I could have done this, but now? I feel like a 9th grader again.
 

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