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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.

College Algebra (MindTap Course List)
College Algebra (MindTap Course List)
12th Edition
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:R. David Gustafson, Jeff Hughes
Chapter3: Functions
Section3.3: More On Functions; Piecewise-defined Functions
Problem 99E: Determine if the statemment is true or false. If the statement is false, then correct it and make it...
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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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