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The linearization of $\ln (1+x)$ at $x=0$ Instead of approximating ln $x$ near $x=1,$ we approximate $\ln (1+x)$ near $x=0$ .We get a simpler formula this way.a. Derive the linearization $\ln (1+x) \approx x$ at $x=0$b. Estimate to five decimal places the error involved in replacing$\ln (1+x)$ by $x$ on the interval $[0,0.1] .$c. Graph $\ln (1+x)$ and $x$ together for $0 \leq x \leq 0.5 .$ Use different colors, if available. At what points does the approximationof $\ln (1+x)$ seem best? Least good? By reading coordinatesfrom the graphs, find as good an upper bound for the error asyour grapher will allow.
a. Use the linearization definition from page 163 .b. 0.00469c. Best on $[0,0.1],$ worst near $0.5 .$ Error is 0.0945348919 (to 10 decimaldigits), upper bound depends on calculator precision.
Calculus 1 / AB
Chapter 7
Transcendental Functions
Section 2
Natural Logarithms
Functions
Missouri State University
Harvey Mudd College
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