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Inverse - difficult to isolate!

  • Thread starter Jet1045
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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 :)
 

Answers and Replies

  • #2
LCKurtz
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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.
 
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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?
 
  • #4
LCKurtz
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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.
 
  • #5
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dx/dy = 1 / (dy/dx) which is easy to calculate once you have the x.
In what cases does that equality hold true?
 
  • #6
LCKurtz
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In what cases does that equality hold true?
Certainly for strictly increasing differentiable functions.
 
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