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Find the incorrect step in the proof and explain why it is incorrect.

Find the incorrect step in the proof and explain why it is incorrect.

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Find the incorrect step in the proof and explain why it is incorrect.

 

Find the error in the following "Proof" and explain in a short paragraph why the reasoning is incorrect. Statement: If f is a function that is differentiable at the point w then the following are equivalent 1. f'(w) = m 2. there is a line y = mx +b containing the point (w, f(w)) and there is &> 0 such that either: f(x) < mx + b for all æ E (w – 8, w+ 8), or, • f(x) > mx +b for all x E (w – 8, w + 8). Proof: To prove that (2) implies (1) begin by observing that since it is known that the line y = mx +b contains the point (w, f(w)) it is possible to find b by evaluating y at w. This gives f(w) = mw + b and hence b = f(w) – mw and so y = mx + f(w) – mw. (5) The other hypothesis is that fis differentiable at w and so f(x) – f(w) lim = f'(w) (6) and f(x) – f(w) lim = f'(w) (7) Two cases must now be considered, Assume first that there is &> O such that f(æ) < mx + b for all æ € (w – , w + 8). It must be shown that f'(w) = m. Note that by Equation (5) f(x) – f(w) < mx + f(w) f(w) = m(x – w). тw Hence, if x > wthen 1/(x – w) >0 and so f(x) – f(w) m(x – w) and therefore by Equation (6) f'(w) = lim f(x) – f(w) m(x – w) < lim = m I - w I - w On the other hand, if a < w then 1/(x – w) <0 and so f(x) – f(w) m(x – w) w and hence by Equation (7) f'(w) = lim f(x) – f(w) m(x – w) > lim I - w It follows that f'(w) = m in this case. Now assume that there is 8> O such that f() > mx + b for all z € (w – 8, w+ 8). It must again be shown that f'(w) = m. Note that by Equation (5) f(x) – f(w) > mæ + f(w) - тw - f(w) — т (х — w). Hence, if æ > w then 1/(x – w) > 0 and so f(x) – f(w) т(т — w) and therefore by Equation (6) f'(w) = lim f(x) – f(w) m(r – w) > lim = m On the other hand, if a < w then 1/(x – w) < 0 and so f(x) – f(w) m(x – w) I - w and hence by Equation (7) f'(w) = lim f(x) – f(w) m(x – w) < lim = m It follows that f'(w) = m in this case too. To see that (1) implies (2) use Equation 3.1.2 of the text. This formula yields that the tangent line to f at w has formula y – f(w) = f'(w)(x – w). Since f'(w) = m it follows that y - f(w) = m(x – w) and so y = mx + (f(w) – w). Since this is a tangent line it touches the graph of f at exactly one point on some interval around w and hence one of the two alternatives of (2) must hold with b = f(w) – w.
Transcribed Image Text:Find the error in the following "Proof" and explain in a short paragraph why the reasoning is incorrect. Statement: If f is a function that is differentiable at the point w then the following are equivalent 1. f'(w) = m 2. there is a line y = mx +b containing the point (w, f(w)) and there is &> 0 such that either: f(x) < mx + b for all æ E (w – 8, w+ 8), or, • f(x) > mx +b for all x E (w – 8, w + 8). Proof: To prove that (2) implies (1) begin by observing that since it is known that the line y = mx +b contains the point (w, f(w)) it is possible to find b by evaluating y at w. This gives f(w) = mw + b and hence b = f(w) – mw and so y = mx + f(w) – mw. (5) The other hypothesis is that fis differentiable at w and so f(x) – f(w) lim = f'(w) (6) and f(x) – f(w) lim = f'(w) (7) Two cases must now be considered, Assume first that there is &> O such that f(æ) < mx + b for all æ € (w – , w + 8). It must be shown that f'(w) = m. Note that by Equation (5) f(x) – f(w) < mx + f(w) f(w) = m(x – w). тw Hence, if x > wthen 1/(x – w) >0 and so f(x) – f(w) m(x – w) and therefore by Equation (6) f'(w) = lim f(x) – f(w) m(x – w) < lim = m I - w I - w On the other hand, if a < w then 1/(x – w) <0 and so f(x) – f(w) m(x – w) w and hence by Equation (7) f'(w) = lim f(x) – f(w) m(x – w) > lim I - w It follows that f'(w) = m in this case. Now assume that there is 8> O such that f() > mx + b for all z € (w – 8, w+ 8). It must again be shown that f'(w) = m. Note that by Equation (5) f(x) – f(w) > mæ + f(w) - тw - f(w) — т (х — w). Hence, if æ > w then 1/(x – w) > 0 and so f(x) – f(w) т(т — w) and therefore by Equation (6) f'(w) = lim f(x) – f(w) m(r – w) > lim = m On the other hand, if a < w then 1/(x – w) < 0 and so f(x) – f(w) m(x – w) I - w and hence by Equation (7) f'(w) = lim f(x) – f(w) m(x – w) < lim = m It follows that f'(w) = m in this case too. To see that (1) implies (2) use Equation 3.1.2 of the text. This formula yields that the tangent line to f at w has formula y – f(w) = f'(w)(x – w). Since f'(w) = m it follows that y - f(w) = m(x – w) and so y = mx + (f(w) – w). Since this is a tangent line it touches the graph of f at exactly one point on some interval around w and hence one of the two alternatives of (2) must hold with b = f(w) – w.
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