# 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)

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 :)

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.

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?

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.

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

In what cases does that equality hold true?