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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

Best Answer 

 #5
avatar+26407 
+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
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6+0 Answers

 #1
avatar+81077 
+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
avatar+91491 
+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
avatar+26407 
+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
avatar+91491 
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
avatar+26407 
+5
Best Answer

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
avatar+91491 
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

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