Explanation Given information: The function: f ( x ) = { c x 2 + 2 x if x < 2 x 3 − c x if x ≥ 2 Calculation: The function is defined piecewise. It is continuous on ( − ∞ , 2 ) and ( 2 , ∞ ) because it’s polynomial on these intervals. For the function to be continuous on ( − ∞ , ∞ ) we need to make it continuous at x = 2 and for that we need the condition on f to be lim x → 2 − f ( x ) = f ( 2 ) = lim x → 2 + f ( x ) We have lim x → 2 − f ( x ) = lim x → 2 −

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ISBN: 9781305420878

Author: Stewart

Publisher: CENGAGE L

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Chapter 2.5, Problem 37E

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To find: value of the constant c for which f is continuous on (−∞,∞) .

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Chapter 2.5, Problem 36E

Chapter 2.5, Problem 38E

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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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Chapter 2 Solutions

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value of the constant c for which f is continuous on ( − ∞ , ∞ ) .

Solution Summary: The author explains how to find the value of the constant c for which f is continuous on (-infty,

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To find: value of the constant c for which f is continuous on (−∞,∞) .

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Chapter 2 Solutions