Derivative of Inverse Function

[James Brady]

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]

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]

There's something fishy in the combination problem statement $\leftrightarrow$ solution 1/7 . James ?

 

[James Brady]

My mistake guys,

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

 

[PeroK]

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]

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

 

[James Brady]

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]

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

 

[BvU]

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

 

[James Brady]

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]

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]

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]

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!)

 

[James Brady]

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.

 

[BvU]

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.