let f(x)= $$\left({\frac{{\mathtt{1}}}{{\mathtt{4}}}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{3}}}\right){\mathtt{\,\small\textbf+\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{1}}$$, solve for (f^-1)'(3).

GIJane43
Feb 22, 2015

#1**+5 **

It looks like you want to find the inverse of f(x)...(* not* the "inverse derivative" )....this one is not easy.....

WolframAlpha gives the inverse as....

To evaluate f^{-1}(3).....we can evaluate f(x) where y= 3

So

3 = (x^3)/4 +x - 1 .... when x =2 .....see this graph......https://www.desmos.com/calculator/qshuoxun0d

So, if (2, 3) is on f(x) then (3, 2) is on the inverse.....thus f-1(3) = 2

CPhill
Feb 23, 2015

#2**+5 **

Well Chris, what you have done looks all very impressive but I'm going to have a go at it myself.

I am treading on shaky ground here so if another mathematicians wants to correct me I shall not be too surprise.

let f(x)= $$\left({\frac{{\mathtt{1}}}{{\mathtt{4}}}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{3}}}\right){\mathtt{\,\small\textbf+\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{1}}$$, solve for (f^-1)'(3).

$$\\y=(0.25)x^3+x-1\\\\

$inverse function$\\\\

x=(0.25)y^3+y-1\\\\

\frac{dx}{dy^{-1}}=0.75y^2+1\\\\

\frac{dy^{-1}}{dx}=\frac{1}{0.75y^2+1}\\\\$$

----------------------------------------------

$$\\x=(0.25)y^3+y-1\\\\

When\;x=3\; \\\\

3=0.25*y^3+y-1\\\\

0=0.25y^3+y-4\\\\$$

http://www.wolframalpha.com/input/?i=0%3D0.25y^3%2By-4

The only real solution to this is y=2

$$\\\frac{dy^{-1}}{dx}=\frac{1}{0.75y^2+1}\\\\

So\; when \;x=3,\;\;y=2\\\\

When\; x=3\;\;\quad\frac{dy^{-1}}{dx}=\frac{1}{0.75*4+1}\\\\

When \;x=3\;\;\quad\frac{dy^{-1}}{dx}=\frac{1}{4}\\\\$$

Melody
Feb 23, 2015

#3**+5 **

The normal interpretation of "inverse function" doesn't mean simply replace the x's by y's and the y's by x's. It means if, say, f(x) = y, then f^{-1}(y) = x.

So, to get the (real) inverse to y = x^{3}/4 + x - 1 we need to solve for x in terms of y (taking the only real solution), * and then* switch x and y values. We get y = the expression given by Chris.

.

Alan
Feb 23, 2015

#4**0 **

You have said this before i think Alan but my answer is the same as Chris's and I have found the derivative as requested.

**Can you give me an example of where my way would give a false answer? **

I am assuming that all valid answers that I find are good but that sometimes my answer might not be valid.

I'll see if I can answer my own question

$$\\y=\sqrt{x}-1\qquad y\ge -1,\;\; x\ge 0\\

my\; way\\

$inverse function$\\

x=\sqrt{y}-1\qquad x\ge -1,\;\;\;y\ge 0 \\

y=(x+1)^2\qquad x\ge -1,\;\;\;y\ge 0 \\$$

**What is wrong with that? How should I do it ??**

Melody
Feb 24, 2015