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What is wrong with the following argument claiming to prove that 1 x² (-/-) There are actually two errors, but you only need to find one. Proof. Let f(x) = 1. Then 1 = = f(x)x. It follows by Theorem 3.2.6 that d dx = d d d 0= -(1) = (ƒ(x)x) = xƒ'(x) + f(x)(x) = xf'(x) + f(x) dx dx dx It follows that xf'(x) = -f(x) and, recalling the definition of f(x), yields that x f'(x) = -1/₁ X

What is wrong with the following argument claiming to prove that 1 x² (-/-) There are actually two errors, but you only need to find one. Proof. Let f(x) = 1. Then 1 = = f(x)x. It follows by Theorem 3.2.6 that d dx = d d d 0= -(1) = (ƒ(x)x) = xƒ'(x) + f(x)(x) = xf'(x) + f(x) dx dx dx It follows that xf'(x) = -f(x) and, recalling the definition of f(x), yields that x f'(x) = -1/₁ X

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter5: Inverse, Exponential, And Logarithmic Functions
Section5.6: Exponential And Logarithmic Equations
Problem 64E
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1. What is wrong with the following argument claiming to prove that 1 x² There are actually two errors, but you only need to find one. Proof. Let f(x) = . Then 1 = f(x)x. It follows by Theorem 3.2.6 that d dx 0 = dx = d #()-- dx Solving for f'(x) yields that d -(ƒ(x)x)= xƒ'(x) + f(x)- ·(x) = xƒ'(x) + f(x) dx It follows that xf'(x) = -f(x) and, recalling the definition of f(x), yields that 1 xf'(x) = − ¹ X d dx X = = f'(x) = == 1
Transcribed Image Text:1. What is wrong with the following argument claiming to prove that 1 x² There are actually two errors, but you only need to find one. Proof. Let f(x) = . Then 1 = f(x)x. It follows by Theorem 3.2.6 that d dx 0 = dx = d #()-- dx Solving for f'(x) yields that d -(ƒ(x)x)= xƒ'(x) + f(x)- ·(x) = xƒ'(x) + f(x) dx It follows that xf'(x) = -f(x) and, recalling the definition of f(x), yields that 1 xf'(x) = − ¹ X d dx X = = f'(x) = == 1
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