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Old 2019-01-25, 14:05   #166
Dr Sardonicus
 
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Quote:
Originally Posted by jvang View Post
<snip>
\(f(x) = x^3+2x+4\), and \(g(x)\) is its inverse. Find \(g(7)\) without finding the formula for \(g(x)\), then find \(g'(7)\).
<snip>
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
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Old 2019-01-25, 17:23   #167
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Quote:
Originally Posted by Dr Sardonicus View Post
...
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.
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Old 2019-01-25, 18:11   #168
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Quote:
Originally Posted by S485122 View Post
(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.
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Old 2019-01-25, 19:02   #169
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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.
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Old 2019-02-08, 02:23   #170
jvang
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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
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Old 2019-02-08, 16:03   #171
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Quote:
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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
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Old 2019-02-13, 03:11   #172
jvang
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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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