Find the first step in the following "Proof" that is incorrect, and explain in a short paragraph why the reasoning is incorrect. Statement: The equation 204 – 50 + 30 = 0ª + 0² – 30 + cos(0) (8) has a positive solution. In other words, there is some positive number 0 satisfying (8). Proof: By dividing both sides of the Equation (8) by o4 + o² – 30 + cos(0) it follows Equation (8) has a solution if 204 – 502 + 30 = 1 (9) 04 - 30 + cos(0) does. Let 204 – 502 + 30 F(0) = (10) 04 + 02 – 30 + cos(0) * From (2.5.4) of the text it follows that cos is a continuous function. By Theorem 2.4.2 (i), (ii) and (iii) it follows that 204 – 50 + 30 and 04 + 502 + 30 + cos(0) are both continuous functions. By applying Theorem-2.4.2 (v) it follows that F is also a continuous function of 0. Then Equation (8) will have a positive solution provided that F(0) = 1 does. This will be established by using Theorem 2.6.1. Begin by noting that if 0 = 1 then 204 – 502 + 30 =0 and hence F(1) = 0. While it is a natural idea that that evaluating 204 – 502 + 30 at 0 = 1 will be useful, finding a value of 0 such that 204 – 502 + 30 > 1 is not that obvious. In such situations it can be useful to evaluate the limit at 0 goes to oo. With this in mind, observe that 204 502 + 30 lim 000 04 + 02 30 + cos(0) 204/04 – 50/04 + 30/04 lim O 100 04/04 + 02/04 – 30/04 + cos(0)/04 2 – 50-2 + 30/0–3 lim O700 1+0-2 – 30-3 + cos(0)/04 + 30/0-3 lim, 00 1+0-2 – 30-3 + cos(0)/04 -2 limo +00 2 – 50 = 2. Let e = 1/2. Then there is some K such that |F(0) – 2| < e= 1/2 for all 0 > K. Let O be greater than both K and 1. Then |F(O) – 2| < 1/2 and so F(O) > 3/2 > 1. To summarize, the following facts about F have been shown: F is continuous at all 0 > 0 F(1) = 0 < 1 there is some O > 1 such that F(O) > 1. It then follows from Theorem 2.6.1 that there must be some 0 between 1 and O such that F(0) = 1.
Find the first step in the following "Proof" that is incorrect, and explain in a short paragraph why the reasoning is incorrect. Statement: The equation 204 – 50 + 30 = 0ª + 0² – 30 + cos(0) (8) has a positive solution. In other words, there is some positive number 0 satisfying (8). Proof: By dividing both sides of the Equation (8) by o4 + o² – 30 + cos(0) it follows Equation (8) has a solution if 204 – 502 + 30 = 1 (9) 04 - 30 + cos(0) does. Let 204 – 502 + 30 F(0) = (10) 04 + 02 – 30 + cos(0) * From (2.5.4) of the text it follows that cos is a continuous function. By Theorem 2.4.2 (i), (ii) and (iii) it follows that 204 – 50 + 30 and 04 + 502 + 30 + cos(0) are both continuous functions. By applying Theorem-2.4.2 (v) it follows that F is also a continuous function of 0. Then Equation (8) will have a positive solution provided that F(0) = 1 does. This will be established by using Theorem 2.6.1. Begin by noting that if 0 = 1 then 204 – 502 + 30 =0 and hence F(1) = 0. While it is a natural idea that that evaluating 204 – 502 + 30 at 0 = 1 will be useful, finding a value of 0 such that 204 – 502 + 30 > 1 is not that obvious. In such situations it can be useful to evaluate the limit at 0 goes to oo. With this in mind, observe that 204 502 + 30 lim 000 04 + 02 30 + cos(0) 204/04 – 50/04 + 30/04 lim O 100 04/04 + 02/04 – 30/04 + cos(0)/04 2 – 50-2 + 30/0–3 lim O700 1+0-2 – 30-3 + cos(0)/04 + 30/0-3 lim, 00 1+0-2 – 30-3 + cos(0)/04 -2 limo +00 2 – 50 = 2. Let e = 1/2. Then there is some K such that |F(0) – 2| < e= 1/2 for all 0 > K. Let O be greater than both K and 1. Then |F(O) – 2| < 1/2 and so F(O) > 3/2 > 1. To summarize, the following facts about F have been shown: F is continuous at all 0 > 0 F(1) = 0 < 1 there is some O > 1 such that F(O) > 1. It then follows from Theorem 2.6.1 that there must be some 0 between 1 and O such that F(0) = 1.
Calculus: Early Transcendentals
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Author:James Stewart
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Chapter1: Functions And Models
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Question
Why is the first step of the proof is incorrect?
Transcribed Image Text:Find the first step in the following "Proof" that is incorrect, and explain in a short paragraph why the reasoning is
incorrect.
Statement: The equation
204 – 50 + 30 = 0ª + 0² – 30 + cos(0)
(8)
has a positive solution. In other words, there is some positive number 0 satisfying (8).
Proof: By dividing both sides of the Equation (8) by
o4 + o² – 30 + cos(0)
it follows Equation (8) has a solution if
204 – 502 + 30
= 1
(9)
04
- 30 + cos(0)
does. Let
204 – 502 + 30
F(0) =
(10)
04 + 02 – 30 + cos(0) *
From (2.5.4) of the text it follows that cos is a continuous function. By Theorem 2.4.2 (i), (ii) and (iii) it follows that
204 – 50 + 30 and 04 + 502 + 30 + cos(0) are both continuous functions. By applying Theorem-2.4.2 (v) it
follows that F is also a continuous function of 0. Then Equation (8) will have a positive solution provided that
F(0) = 1 does.
This will be established by using Theorem 2.6.1. Begin by noting that if 0 = 1 then 204 – 502 + 30 =0 and hence
F(1) = 0. While it is a natural idea that that evaluating 204 – 502 + 30 at 0 = 1 will be useful, finding a value of 0
such that 204 – 502 + 30 > 1 is not that obvious. In such situations it can be useful to evaluate the limit at 0 goes
to oo. With this in mind, observe that
204
502 + 30
lim
000 04
+ 02
30 + cos(0)
204/04 – 50/04 + 30/04
lim
O 100 04/04 + 02/04 – 30/04 + cos(0)/04
2 – 50-2 + 30/0–3
lim
O700 1+0-2 – 30-3 + cos(0)/04
+ 30/0-3
lim, 00 1+0-2 – 30-3 + cos(0)/04
-2
limo +00 2 – 50
= 2.
Let e = 1/2. Then there is some K such that |F(0) – 2| < e= 1/2 for all 0 > K. Let O be greater than both K
and 1. Then |F(O) – 2| < 1/2 and so F(O) > 3/2 > 1. To summarize, the following facts about F have been
shown:
F is continuous at all 0 > 0
F(1) = 0 < 1
there is some O > 1 such that F(O) > 1.
It then follows from Theorem 2.6.1 that there must be some 0 between 1 and O such that F(0) = 1.
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