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).
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
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
$$\\\frac{dy^{-1}}{dx}=\frac{1}{0.75y^2+1}\\\\
So\; when \;x=3,\;\;y=2\\\\
When\; x=3\;\;\quad\frac{dy^{-1}}{dx}=\frac{1}{0.75*4+1}\\\\
When \;x=3\;\;\quad\frac{dy^{-1}}{dx}=\frac{1}{4}\\\\$$
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.
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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 ??