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2019-01-25, 14:05   #166
Dr Sardonicus

Feb 2017
Nowhere

23×223 Posts

Quote:
 Originally Posted by jvang $$f(x) = x^3+2x+4$$, and $$g(x)$$ is its inverse. Find $$g(7)$$ without finding the formula for $$g(x)$$, then find $$g'(7)$$.
f(x) = x3 + 2*x + 4

y = x3 + 2*x + 4, point (1, 7)

y' = 3*x2 + 2 At (1, 7) y' = 5

Inverse function

x = y3 + 2*y + 4, point (7, 1)

1 = (3*y2 + 2)*y' At (7, 1) y' = 1/5

Using the "just transpose x and y" idea, the point-slope 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 point-slope form,

(1/5)*(x - 1) = y - 1.

Last fiddled with by Dr Sardonicus on 2019-01-25 at 14:11

2019-01-25, 17:23   #167
S485122

"Jacob"
Sep 2006
Brussels, Belgium

2·877 Posts

Quote:
 Originally Posted by Dr Sardonicus ... Inverse function x = y3 + 2*y + 4, point (7, 1) ...
IMHO opinion your definition of the inverse function is not correct.
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=y3 is not the inverse function of y=x3, since substituting x for your inverse function will give y=(y3)3=x9.)

Jacob

See Wikipedia : Inverse function for instance.

2019-01-25, 18:11   #168
VBCurtis

"Curtis"
Feb 2005
Riverside, CA

37×137 Posts

Quote:
 Originally Posted by S485122 (For instance it is obvious that x=y3 is not the inverse function of y=x3, since substituting x for your inverse function will give y=(y3)3=x9.) Jacob
If you rewrite x = y3 to be in the format of y as a function of x, then the definition you cite applies. You're splitting hairs to say that y = x1/3 is the inverse, but x = y3 is not.

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.

2019-01-25, 19:02   #169
Dr Sardonicus

Feb 2017
Nowhere

23×223 Posts

Quote:
 Originally Posted by S485122 IMHO opinion your definition of the inverse function is not correct. 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=y3 is not the inverse function of y=x3, since substituting x for your inverse function will give y=(y3)3=x9.)
Your substitution is inconsistent with the definition you give. Using the example y = x^3, the inverse function implicitly defined by x = y^3 is not y^3, but rather g(x) = y. And, by its formulation (g(x))3 = x.

The only real difficulty with the implicit formulation of the inverse function is that it may not be well-defined. And even then, the problem only really manifests itself at points where two or more possible inverses meet.

 2019-02-08, 02:23 #170 jvang veganjoy     "Joey" Nov 2015 Middle of Nowhere,AR 22·3·37 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: $$y = x^3-2x^2+5x$$ 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
2019-02-08, 16:03   #171
Dr Sardonicus

Feb 2017
Nowhere

140916 Posts

Quote:
 Originally Posted by jvang I'm trying to figure out the inverse of this, so that I can find the derivative of the inverse: $$y = x^3-2x^2+5x$$ or something very similar. You just swap the variables and solve for y, right?
You went one step too far. Don't try to "solve for y," just swap the variables, then use implicit differentiation.

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 2019-02-08 at 16:39

 2019-02-13, 03:11 #172 jvang veganjoy     "Joey" Nov 2015 Middle of Nowhere,AR 22·3·37 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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