+0

# Inverse derivatives

0
228
6
+11

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

#5
+26399
+5

Yes, ok Melody, but in your original answer you didn't really give the inverse function, which would be y as a function of x (and if there had been more than one real answer, which would you have chosen?).

Alan  Feb 24, 2015
Sort:

#1
+80983
+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
+91446
+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
+26399
+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 = x3/4 + 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
+91446
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
#5
+26399
+5

Yes, ok Melody, but in your original answer you didn't really give the inverse function, which would be y as a function of x (and if there had been more than one real answer, which would you have chosen?).

Alan  Feb 24, 2015
#6
+91446
0

Thank you Alan,

I still don't understand but it doesn't matter.  I don't understand anything much lately.

Thank you for trying to explain.

Melody  Feb 24, 2015

### 10 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details