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Inverse function help

  • Thread starter aquitaine
  • Start date
  • #1
18
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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?
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,772
911
(a+ b)3 is NOT a3+ b3

It is a3+ 3a2b+ 3ab2+ b3
 
  • #3
djeitnstine
Gold Member
614
0
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?
Your equation is a tad confusing, you have y = [tex]\sqrt[3]{x+sqrt(1+x^2)}[/tex] + [tex]\sqrt[3]{x-sqrt(1+x^2)}
[/tex]

then you cube both sides and switch y's and x's to get [tex]x^{3}[/tex] = y+[tex]\sqrt(1+y^2)[/tex] + y-[tex]\sqrt(1+y^2)[/tex]

Which is wrong. Now if your equation was y = [tex]\sqrt[3]{x+\sqrt(1+x^2)+x-\sqrt(1+x^2)}[/tex] this method would be correct. However, it would have been simplified easily before you even cube both sides; In this case it would be in the form y = [tex]\sqrt[3]{2x}[/tex] for the roots cancel automatically.
 
Last edited:

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