# Inverse - difficult to isolate!

## 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)

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
Homework Helper
Gold Member
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.

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
Homework Helper
Gold Member
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.

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

In what cases does that equality hold true?

LCKurtz
Homework Helper
Gold Member
In what cases does that equality hold true?

Certainly for strictly increasing differentiable functions.