20190125, 14:05  #166  
Feb 2017
Nowhere
5·997 Posts 
Quote:
y = x^{3} + 2*x + 4, point (1, 7) y' = 3*x^{2} + 2 At (1, 7) y' = 5 Inverse function x = y^{3} + 2*y + 4, point (7, 1) 1 = (3*y^{2} + 2)*y' At (7, 1) y' = 1/5 Using the "just transpose x and y" idea, the pointslope equation for the tangent line to y = f(x) at (1, 7) is y  7 = 5*(x  1). An equation for the tangent line to x = f(y) at (7, 1) is then x  7 = 5*(y  1). Casting this into pointslope form, (1/5)*(x  1) = y  1. Last fiddled with by Dr Sardonicus on 20190125 at 14:11 

20190125, 17:23  #167  
"Jacob"
Sep 2006
Brussels, Belgium
6CF_{16} Posts 
Quote:
If y=f(x) the inverse function y=f^{ 1}(x) will be such that f(g(x))=y this is obviously not the case your calculations. (For instance it is obvious that x=y^{3} is not the inverse function of y=x^{3}, since substituting x for your inverse function will give y=(y^{3})^{3}=x^{9}.) Jacob See Wikipedia : Inverse function for instance. 

20190125, 18:11  #168  
"Curtis"
Feb 2005
Riverside, CA
2^{2}×1,249 Posts 
Quote:
To find an inverse, switch x and y. If you want the inverse in g(x) format, then solve for y after switching x and y. 

20190125, 19:02  #169  
Feb 2017
Nowhere
5×997 Posts 
Quote:
The only real difficulty with the implicit formulation of the inverse function is that it may not be welldefined. And even then, the problem only really manifests itself at points where two or more possible inverses meet. 

20190208, 02:23  #170 
veganjoy
"Joey"
Nov 2015
Middle of Nowhere,AR
2^{2}·3·37 Posts 
There was a problem we were working on today, in which we're still doing inverse derivative stuff. I'm trying to figure out the inverse of this, so that I can find the derivative of the inverse: \(y = x^32x^2+5x\) or something very similar. You just swap the variables and solve for y, right?\[x = y^3  2y^2 +5y\]But where do you go from here? I can't figure out how to isolate the y, since any sort of division leaves an extra y in a denominator

20190208, 16:03  #171  
Feb 2017
Nowhere
4985_{10} Posts 
Quote:
Hmm. The cubic x^3  2*x^2 + 5*x  a has discriminant 27*a^2 + 148*a  400, which is negative for any real a. So Cardano's formulas will give the real solution to x^3  2*x^2 + 5*x = a as a sum of a rational number and two real cube roots, but it will be an algebraic mess. Differentiating would produce an even bigger mess. Besides  implicit differentiation works just fine, even if there is no "nice" formula for the inverse function. Last fiddled with by Dr Sardonicus on 20190208 at 16:39 

20190213, 03:11  #172 
veganjoy
"Joey"
Nov 2015
Middle of Nowhere,AR
2^{2}·3·37 Posts 
We covered a bit on matching the derivatives of graphs with their originals. One thing that was interesting was inflection points, which has something to do with concavity (which we haven't covered, also something about "critical numbers"?). From my observations of derivative graphs, it seems that inflection points are located at the x intercepts of the original graph's second derivative, which would be where the slope of the slope of the graph is 0... is there a better way to word that?

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