1. Consider the curve C = {(x, y) € R²: x – y³ = 0}. Let ø: R → R be defined by (t) = t³, then o is a bijection from R to R and so it has an inverse o-1. The equation that defines C can be written as a = ¢(y), and by applying o-1 on both sides we obtain the equivalent equation y = 0-'(x). Therefore y is a function of r. Taking this into account, differentiating the equation y x = y° with respect to x on both sides yields 1 = 3y2 and y = 0 into the last equation yields 1 = 0. fip dr Since (0, 0) e C, plugging in r = 0 (a) Explain the mistake(s) in the above argument. (b) Explain why the Implicit Function Theorem cannot be used to fix the above argument.

Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter8: Polynomials
Section8.5: Solution Of Cubic And Quartic Equations By Formulas (optional)
Problem 29E
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Solve questions 1a and b 

1. Consider the curve C = {(x, y) E R²: x – y3 = 0}. Let o: R → R be defined by ø(t) = t³,
then o is a bijection from R to R and so it has an inverse o-1. The equation that defines C can
be written as x = ¢(y), and by applying o-1 on both sides we obtain the equivalent equation
y = 0-1(x). Therefore y is a function of x. Taking this into account, differentiating the equation
y with respect to x on both sides yields 1 = 3y
da
dy
Since (0,0) e C, plugging in x = 0
and y = 0 into the last equation yields 1 = 0.
(a) Explain the mistake(s) in the above argument.
(b) Explain why the Implicit Function Theorem cannot be used to fix the above argument.
Transcribed Image Text:1. Consider the curve C = {(x, y) E R²: x – y3 = 0}. Let o: R → R be defined by ø(t) = t³, then o is a bijection from R to R and so it has an inverse o-1. The equation that defines C can be written as x = ¢(y), and by applying o-1 on both sides we obtain the equivalent equation y = 0-1(x). Therefore y is a function of x. Taking this into account, differentiating the equation y with respect to x on both sides yields 1 = 3y da dy Since (0,0) e C, plugging in x = 0 and y = 0 into the last equation yields 1 = 0. (a) Explain the mistake(s) in the above argument. (b) Explain why the Implicit Function Theorem cannot be used to fix the above argument.
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