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Is this a possible way to show that f(x) = arctan(2(x-1)) - lnIxI has exactly four real roots?

Is this a possible way to show that f(x) = arctan(2(x-1)) - lnIxI has exactly four real roots?

Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
6th Edition
ISBN:9781337111348
Author:Bruce Crauder, Benny Evans, Alan Noell
Publisher:Bruce Crauder, Benny Evans, Alan Noell
Chapter2: Graphical And Tabular Analysis
Section2.1: Tables And Trends
Problem 1TU: If a coffee filter is dropped, its velocity after t seconds is given by v(t)=4(10.0003t) feet per...
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Is this a possible way to show that f(x) = arctan(2(x-1)) - lnIxI has exactly four real roots?

Prove that fex)= arctan (2 (x-1)) - In 1x1 has 4 rocts. First we prove that we cannot have more than four roots. By contradiction, let's assume we have five or mavo roots. By Rolle's Thm. the first derivative would have to have four or more roots. The derivative is -4x² +10x-5 and does not have more than two roots, a camad.! 4x³8x² +5x 2 Now, we will prove that we do have four roots. We will use the IVT (Bolzano's Thm). We will start substituting nice numbers in f, waiting for the value to change signs. ·2|-1|-0.25 0.25/ 0.5 | 1.25|2|3|4|5 X-2-1025/0 f(x) Sign Intervals: (-1,-0.25), (0.25, 0.5), (0.5, 4), (4, 5) Hence, fix) has exactly four real roots! -2....-1.32 0.196 0.403 -0.09.. 0.24... 0.4 6.227.0.02 -0... + + + +++ - 1
Transcribed Image Text:Prove that fex)= arctan (2 (x-1)) - In 1x1 has 4 rocts. First we prove that we cannot have more than four roots. By contradiction, let's assume we have five or mavo roots. By Rolle's Thm. the first derivative would have to have four or more roots. The derivative is -4x² +10x-5 and does not have more than two roots, a camad.! 4x³8x² +5x 2 Now, we will prove that we do have four roots. We will use the IVT (Bolzano's Thm). We will start substituting nice numbers in f, waiting for the value to change signs. ·2|-1|-0.25 0.25/ 0.5 | 1.25|2|3|4|5 X-2-1025/0 f(x) Sign Intervals: (-1,-0.25), (0.25, 0.5), (0.5, 4), (4, 5) Hence, fix) has exactly four real roots! -2....-1.32 0.196 0.403 -0.09.. 0.24... 0.4 6.227.0.02 -0... + + + +++ - 1
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