inverse function

[rubing]

I am trying to find the inverse function of the following equation:

x = y⁴ + y² + 1 , y >= 0

Can you show me how? thanks!!

 

[willmoore21]

Can you find an inverse to a simpler function such as x=y+1?

 

[rubing]

of course, that's easy you just subtract 1 from both sides, which yields:

y = x - 1

 

[HallsofIvy]

Let \(\displaystyle u= y^2\) and you have the quadratic function \(\displaystyle x= u^2+ u+ 1\) which you can treat as the quadratic equation \(\displaystyle u^2+ u+ (1- x)= 0\). Solve it usuing the quadratic formula, then take the square root to get y. Don't forget the y is positive here.

 

[dwsingrs]

I am trying to find the inverse function of the following equation:

x = y⁴ + y² + 1 , y >= 0

Can you show me how? thanks!!

Greetings. I know enough to be dangerous. (I need to work on my Domain & Range competence.) Here goes.

If I correctly understand, the below can be treated as a quadratic.

y⁴ + y2 + 1 = x

y⁴ + y² = x -1

Now, "complete the square" of y⁴ + y²:

y⁴ + y² + (1/2)² = x - 1 + (1/2)² (The 1/2 being squared is the b/2a derived from the quadratic general form y = ax² + bx + c.)

(y² + 1/2)² = x - 3/4

y² + 1/2 = +or- (x - 3/4)^1/2

y² = - 1/2 +or- (x - 3/4)^1/2

y = +or- [- 1/2 +or- (x - 3/4)^1/2 ]^1/2

(Four equations reflected above. By my reckoning, D: x>1 in order to accomplish required R: y>0.)


What I myself am having trouble with (and will shortly post) is calculating the inverse of the likes of y = x² + x^1/2, which is not a quadratic, as opposed to y = x + x^1/2 which, it appears, can be treated as a quadratic.

Cheers!