Explanation Given information: The given inequality implication is if 0 < | x + 2 | < 0.075 then | x 2 − 4 | < 0.4 . Concept used: For the limit statement lim x → c f ( x ) = L , the delta-epsilon statement is For all ∈ > 0 , there exist δ > 0 such that whenever x ∈ ( c − δ , c ) ∪ ( c , c + δ ) guarantees f ( x ) ∈ ( L − ∈ , L + ∈ ) . Mathematically it can be written as for every x satisfying 0 < | x − c | < δ , every f ( x ) satisfies | f ( x ) − L | < ∈ . Calculation: Consider the given inequality. 0 < | x + 2 | < 0

14th Edition

ISBN: 9781464151088

Author: Associate Professor Laura Taalman, Peter Kohn

Publisher: W.H. Freeman & Company

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Chapter 1.3, Problem 17C

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Whether the given inequality implication is true or false. Give a counter example for false implication.

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Chapter 1.3, Problem 16C

Chapter 1.3, Problem 18C

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Taylor Series Approximation Example- H.W More terms used implies better approximation f(x) 4 f(x) Zero order f(x + 1) = f(x;) First order f(x; + 1) = f(x;) + f'(x;)h 1.0 Second order 0.5 True f(x + 1) = f(x) + f'(x)h + ƒ"(x;) h2 2! f(x+1) 0 x; = 0 x+1 = 1 x h f(x)=0.1x4-0.15x³- 0.5x2 -0.25x + 1.2 51 Taylor Series Approximation H.w: Smaller step size implies smaller error Errors f(x) + f(x,) Zero order f(x,+ 1) = f(x) First order 1.0 0.5 Reduced step size Second order True f(x + 1) = f(x) + f'(x)h f(x; + 1) = f(x) + f'(x)h + "(xi) h2 f(x,+1) O x₁ = 0 x+1=1 Using Taylor Series Expansion estimate f(1.35) with x0 =0.75 with 5 iterations (or & s= 5%) for f(x)=0.1x 0.15x³-0.5x²- 0.25x + 1.2 52

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Whether the given inequality implication is true or false. Give a counter example for false implication.

Solution Summary: The author explains that the given inequality implication is true or false, and gives a counter example.

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Chapter 1 Solutions