# Derivative of Inverse Function

## 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.

Simon Bridge
Homework Helper
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?

BvU
Homework Helper
There's something fishy in the combination problem statement ##\leftrightarrow ## solution 1/7 . James ?

• PeroK
My mistake guys,

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

PeroK
Homework Helper
Gold Member
2020 Award
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})'##.

BvU
Homework Helper
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 !

 perok: he had a ##\ ## ' ##\ ## missing in
##(f^{-1}(x))=\frac{1}{f'(f^{-1}(x))}## ; should've been $$(f^{-1}(x))'=\frac{1}{f'(f^{-1}(x))}$$

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.

BvU
Homework Helper
Again: you don't want to evaluate the derivative at the point x = 4 but at the point f-1(4)

BvU
Homework Helper
Do you understand the relationship $$(f^{-1}(x))'=\frac{1}{f'(f^{-1}(x))}\ \ \rm ?$$

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.

PeroK
Homework Helper
Gold Member
2020 Award
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!

• BvU
Homework Helper
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))## !

• Ray Vickson
Homework Helper
Dearly Missed
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!)

• 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.

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