Inverse - difficult to isolate

[Jet1045]

Homework Statement

Hello! I am having troubles with a question I just got in my AP calc class. It is:

Let f(x) = x⁵ + 2x³ + x + 1

a) find f ^-1(3) and (f ^-1)'(3)

Homework Equations

N/A

The Attempt at a Solution

Okay so my first idea was obviously to switch the x and y in f(x) giving me x = y⁵ + 2y³ + y + 1 however, it is clearly too difficult to isolate y in this equation.

So i was thinking, for f ^-1(3) , if I substitute 3 for y in the original equation, giving me 3 = x⁵ + 2x³ + x + 1 and solve for the zeroes, will that give me my answer to f ^-1(3) ? And then for (f ^-1)'(3), i would set the derivative of f(x) to 3, and solve for the zeroes, which would look like 3 = 5x⁴ + 6x² + 1

Would doing these things with f(x) effectively give me the answers to these two inverse problems?

Any help would be appreciated :)

 

[LCKurtz]

I think this problem must have a typo in it. If you could find the x where f(x) = 3, yes, that would give you f^-1(3). But I don't think you can do that in any elementary way. There is a long way to solve cubics, but I doubt you have seen that. That's why I think there is a typo if this came up in an AP calculus class.

As for the derivative of the inverse at 3, you probably would use dy/dx = 1 / (dx/dy) if you knew f^-1(3).

Ask your teacher whether the problem is stated as he meant it to be.

 

[Jet1045]

Yeah it could very well be a typo! But he told us that we could use our calculators to find the zeroes of any equations that we encounter... so If i use my calculator, would the zeroes I find be the correct answers?

 

[LCKurtz]

Yeah it could very well be a typo! But he told us that we could use our calculators to find the zeroes of any equations that we encounter... so If i use my calculator, would the zeroes I find be the correct answers?

Ahhh, well that is a horse of a different color. I think you will find if you graph that function that it is increasing and only has one value for x where f(x) = 3, which you can solve approximately numerically. Then use the equation

dx/dy = 1 / (dy/dx) which is easy to calculate once you have the x.

 

[Amok]

dx/dy = 1 / (dy/dx) which is easy to calculate once you have the x.

In what cases does that equality hold true?

 

[LCKurtz]

In what cases does that equality hold true?

Certainly for strictly increasing differentiable functions.

 

[Amok]

Actually it holds as longs as the function has an inverse and dy/dx different than 0 :p

http://en.wikipedia.org/wiki/Inverse_function_theorem

 

[LCKurtz]

Actually it holds as longs as the function has an inverse and dy/dx different than 0 :p

http://en.wikipedia.org/wiki/Inverse_function_theorem

And your point is?

 

[Amok]

And your point is?

Wiki is my friend :)