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# Inverse derivatives

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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

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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
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#1
+85753
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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
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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
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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
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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
+26638
+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
+92217
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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

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