Derivative of Inverse Function

  • #1
103
4

Homework Statement


Suppose ##f(x) = x^5 + 2x + 1## and ##f^{-1}## is the inverse of function f. Evaluate ##f^{-1}(4)##

solution: 1/7

Homework Equations


##(f^{-1}(x))=\frac{1}{f'(f^{-1}(x))}##

The Attempt at a Solution


I attempted to use my calculator's solve function to get the solution of ##4 = 5x^4 + 2##, so the value of x when the derivative is equal to 4. I then took the inverse of this. Sort of still grasping in the dark right now tho.
 

Answers and Replies

  • #2
Simon Bridge
Science Advisor
Homework Helper
17,873
1,655
Your problem statement says you have to:
... Evaluate ##f^{-1}(4)##
Note: ##f^{-1}(f(x))=x \implies f^{-1}(4) = x:f(x)=4##
Why were you computing a derivative?
 
  • #3
BvU
Science Advisor
Homework Helper
14,448
3,737
There's something fishy in the combination problem statement ##\leftrightarrow ## solution 1/7 . James ?
 
  • #4
103
4
My mistake guys,

The problem statement asks for ##(f^{-1})^{'}(4)##
 
  • #5
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
17,537
9,267
My mistake guys,

The problem statement asks for ##(f^{-1})^{'}(4)##

The relevant equation you posted is not that useful, as it involves ##f'##. You need something similar involving ##(f^{-1})'##.
 
  • #6
BvU
Science Advisor
Homework Helper
14,448
3,737
So you don't want to solve for 'derivative = 4' but you want to evaluate the derivative at the point where the function value is 4 !

[edit] perok: he had a ##\ ## ' ##\ ## missing in
##(f^{-1}(x))=\frac{1}{f'(f^{-1}(x))}## ; should've been $$
(f^{-1}(x))'=\frac{1}{f'(f^{-1}(x))}$$
 
  • #7
103
4
Hmmm. ##f^{'} = 5x^4+2## and ##f^{'}(4)= 1282##

So I have the derivative of of the function itself at a certain x value. But I'm not sure how I can translate that into finding the derivative of the inverse at the same x value.
 
  • #8
BvU
Science Advisor
Homework Helper
14,448
3,737
Again: you don't want to evaluate the derivative at the point x = 4 but at the point f-1(4)
 
  • #9
BvU
Science Advisor
Homework Helper
14,448
3,737
Do you understand the relationship $$
(f^{-1}(x))'=\frac{1}{f'(f^{-1}(x))}\ \ \rm ?$$
 
  • #10
103
4
I understand this relationship, but I do not know how I can evaluate ##f^{'}(4)## since I can't figure out how to get the inverse of the original function. Plugging in for ##y=x^5 + 2x + 1##, I can't figure out how I would be able to solve for x in terms of y.
 
  • #11
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
17,537
9,267
I understand this relationship, but I do not know how I can evaluate ##f^{'}(4)## since I can't figure out how to get the inverse of the original function. Plugging in for ##y=x^5 + 2x + 1##, I can't figure out how I would be able to solve for x in terms of y.

You could always just take a guess!
 
  • Like
Likes James Brady
  • #12
BvU
Science Advisor
Homework Helper
14,448
3,737
I understand this relationship, but I do not know how I can evaluate ##f^{'}(4)## since I can't figure out how to get the inverse of the original function. Plugging in for ##y=x^5 + 2x + 1##, I can't figure out how I would be able to solve for x in terms of y.
Try a few numbers, or do bisection ( f(0) = 1, f(2) >> 4 so...)

And again (again) :
but I do not know how I can evaluate f′(4)
You don't want that. You want ##f'(f^{-1}(4))## !
 
  • Like
Likes James Brady
  • #13
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722
I understand this relationship, but I do not know how I can evaluate ##f^{'}(4)## since I can't figure out how to get the inverse of the original function. Plugging in for ##y=x^5 + 2x + 1##, I can't figure out how I would be able to solve for x in terms of y.
Plot the graph of ##y = x^5+2x+1## over some modest ##x##-range; that will give you a good feeling for approximately where you have the solution of ##f(x)=4.## Alternately you can try using the "rational root theorem" on the equation ##x^5 + 2x - 3 = 0.## (Once you get the solution you will wonder why you did not see it right away!)
 
  • Like
Likes James Brady
  • #14
103
4
You don't want that. You want f′(f−1(4))f'(f^{-1}(4)) !

Thanks guys, I guess I just really needed to sleep on it, judging by my sloppiness. Coming back today, I can see where you all were going:

##(f^{-1}(x))^{'} = \frac{1}{f^{'}(f^{-1}(x))}##

First thing, we solve the interior function on the RHS, or ##4 = x^5 + 2x +1##. I used the numerical solver on the calculator, but I see now that 1 could be obvious if you take a step back for a second. After that, we're no longer working with inverses, the derivative function ##f^{'}(x) = 5x^4 + 2## taken at x = 1. One over that and boom.

Thanks again for your patience.
 
  • #15
BvU
Science Advisor
Homework Helper
14,448
3,737
I guess I just really needed to sleep on it
Important learning experience: brains are a lot like muscles, they need to recover from time to time or else things go sour. Sometimes dropping your pencil and doing some R&R is the quickest way to achieve some progress when stuck with a problem.
 

Related Threads on Derivative of Inverse Function

  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
7
Views
801
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
12
Views
12K
  • Last Post
Replies
3
Views
2K
Replies
15
Views
2K
Replies
3
Views
2K
Top